国人过于高看诺贝尔奖了
Originally published at 狗和留美者不得入内. You can comment here or there.
今天在已发表至少一个月而得到不少赞同的此文章改了个错别字，完了却被删除了，我把内
今天的诺贝尔奖的科研工作意义显然远不如前，因为基础科学已经遇到瓶颈了。今天的科研
那么多不了解科研怎么回事儿的国人把诺贝尔奖获得者似乎视为万能的神，这是搞笑的。发
就不用说今天随着西方衰落的趋势，西方人包括西方名校教授的评价越来越无关了。中国的
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Originally published at 狗和留美者不得入内. You can comment here or there.
今天在已发表至少一个月而得到不少赞同的此文章改了个错别字，完了却被删除了，我把内
今天的诺贝尔奖的科研工作意义显然远不如前，因为基础科学已经遇到瓶颈了。今天的科研
那么多不了解科研怎么回事儿的国人把诺贝尔奖获得者似乎视为万能的神，这是搞笑的。发
就不用说今天随着西方衰落的趋势，西方人包括西方名校教授的评价越来越无关了。中国的
Originally published at 狗和留美者不得入内. You can comment here or there.
Let be a vector space over endowed with an inner product . Assume that it has an orthonormal basis . This means any can be expressed as
where for all . For each , we take any arbitrary sequence of reals that converges to . Let . We can also require that for all and that , which means uniform convergence across all . If we want that a cofinite number of the sequences at index are at least a certain distance away from the value converged to, we additionally define and require that for all , . Obviously, for all . Setting obviously satisfies these requirements.
We notice that iff the associated series converges, which of course does not happen all the time, which means that that via is not well defined. Thus, the two-norm on finite dimensional vector spaces induced by the inner product cannot be defined similarly on infinite dimensional vector spaces.
Proposition 1 There exists an infinite dimensional normed vector space that is a Banach space.
Proof: Keep in mind that for all , we must have one of the properties of norm. For simplicity we shall require that for any such that ,
which is of course compatible with triangle inequality. We shall require that is a Banach space which means that if is convergent, it is in . Moreover, we require that if converges, then . We now show that this is well-defined. We note that by ,
Assume that the infinite series converges, by definition, for any ,
Then, that
implies that
Applying absolute homogeneity of the norm to gives us
If is convergent, then must of course be bounded based on our extension of norm to the infinite series. Assume it converges to . Then by , . Thus, we have
which verifies absolute homogeneity. Verifying triangle inequality is straightforward and we leave this to the reader.
For , it is necessary that for an arbitrary real sequence . By the aforementioned properties, if we prescribe values for for all . If we require that for any , the associated coefficients satisfy , then values for such that suffices. It is easy to see that for any sequence of absolutely bounded real coefficients, there is some upper bound , which implies that the norm of the vector in associated with it is upper bounded by .
Finally, one easily verifies closure under scalar multiplication and addition of this vector space, wherein the vector coefficients are guaranteed to have bounded supremum.
We note that the bounded supremum requirement is the more essential part here. There is a in fact simple stupid way to construct a norm which is given by . In order for this to be a norm, we must of course always have .
Example 1 Let . Then, .
Example 2 We can let .
Example 3 We can define a sequence of vectors converging to via sequences of coefficients , we indeed have with respect to this norm. More specifically, goes to as . Thus, sequence converges with respect to norm given in Example 2. It also does with respect to the norm given in Example 1.
Example 4 If we let for all , then we would need to restrict the vector space elements to correspond to sequences such that .
Example 5 We let . Using the norm given in Example 1, we have that for all , . The tells us that it also cannot converge to any value in other . Thus does not converge with respect to the Example 1 norm. It does converge to with respect to the Example 2 norm though.
Proposition 2 Any is uniquely defined by prescribing the values of over all . In order for to be well-defined, we must have .
Proof: Follows directly from definition of basis and linearity of .
Definition 1 We say that , or that converges to strongly with respect to some norm iff .
Definition 2 We say that , or that converges to weakly, iff for all , .
Proposition 3 The sequence in Example 5 converges weakly (in the Example 1 norm).
Proof: If , then by Proposition 2, we have . Let and , with of course monotonically from above.
Then, we have that . That this upper bound converges to from above completes our proof.
Definition 3 The coarsest topology on normed vector space with respect to which every is continuous, which we shall denote with , is the weak topology.
Proposition 4 iff every neighborhood of with respect to contains for some all such that .
Proof: Follows directly from the definition of and of neighborhood.
Proposition 5 is a neighborhood basis of . If we let denote an arbitrary finite subset of , then the every set of the aforementioned collection can be denoted as .
Proof: Satisfaction of the continuity requirement means that for any open interval in , for any . We thus take to be a subbase of . (For the definition of subbase, see Definition 1 of [2].) Every open interval in has a midpoint which we take to be , and we take to be half the interval length. The collection of sets given in the proposition are all finite intersections of elements of . Every is the arbitrary union of finite intersections of elements of , by definition of subbase. Thus, either is the empty set, or it contains some . This completes our proof.
Definition 4 The strong topology of a normed vector space is the coarsest topology such that the norm is a continuous function.
The norm on the dual space is defined as in Definition 8 of [3]. This norm gives rise to a strong topology on the dual space as well.
Lemma 1 For any , is closed in the strong topology on .
Proof: is a closed set and the norm is a continuous function. Thus, is also closed. Proposition 7 of [4] tells us that the preimage of a closed set, when the function is continuous is also closed.
Proposition 6 For any , .
Proof: We note that if not, then would not be a well-defined norm on . By definition of , is continuous. By Proposition 2 of [3], continuous implies bounded. The final equality in Definition 8 of [3] then gives us the desired result.
Proposition 7 For any normal vector space , the weak topology is a subset of the strong topology and strong convergence implies weak convergence.
Proof: It suffices to prove that strong convergence implies weak convergence. Assume . Then, by Proposition 6, for any , , with , which completes our proof.
References
Originally published at 狗和留美者不得入内. You can comment here or there.
We shall use n.v.s to refer to normed vector space.
Definition 1 Let be a n.v.s. A set of the form is called a hyperplane.
Definition 2 We call a set of the form a half-space determined by . Replacing with gives the other half-space.
Proposition 1 Any half-space is a convex set.
Proof: Trivial.
Proposition 2 Let be a n.v.s. such that . For any , the hyperplane is closed iff is continuous.
Proof: open iff is closed. Suppose is continuous and that for , . Take some open neighborhood of not containing . Then, is open and disjoint with .
To prove that is continuous, it suffices to prove that is bounded, by Proposition 2 of [1]. To show that is bounded, showing that suffices, where is the unit ball centered at zero. To do so, it suffices to show that every such that has an open neighborhood of the form such that for any , , since if this holds, then for ,
Thus, .
Assume that is open. Then, every such that is contained by a . Suppose that satisfies . Then, there must exist some such that , a contradiction. This completes our proof.
Definition 3 Let be two subsets of and with . If is such that of its two half-planes, one contains and the other contains , then we say that separates and . We say that strictly separates and if there exists some such that
We now wish to prove that for any open convex set containing and any , there exists a hyperplane that separates and . Let denote the linear functional associated with this separation. We prescribe for some . If we can find some sublinear function that is strictly bounded above on by such that on the subspace satisfies , then we can apply the Hahn-Banach theorem to extend to all in order separate from by a hyperplane.
For this desired sublinear function , we try the following.
Definition 4 Let be a subset of n.v.s . Then the gauge or Minkowski functional with respect to is defined by .
Proposition 3 If contains an open ball centered at , the Minkowski functional satisfies the condition that for all , .
Proof: Trivial and left to the reader.
We want that on , , which is satisfied when we let . We let . That is open means there is some open ball centered at the origin . For any , obviously holds. This is a simple stupid way to uniformly upper bound on .
Proposition 4 For any subset , is bounded above by for . If is open, then for all , .
Proof: Trivially proven.
Proposition 5 If is an open convex set containing , then .
Proof: Since we can scale down arbitrarily by Proposition 3, it suffices to prove this triangle inequality on an open ball centered at that is contained by . Moreover it suffices to prove that for any ,
is satisfied. We notice that
with
Let . Obviously . With this in mind, that is convex implies that for any for any , . We use to set the value of noticing that we can multiply by the denominator to derive the desired inequality. We calculate
We notice that if we replace with , we would obtain the desired inequality. Because is convex must be connected. Assuming that , we have . The connectedness hypothesis implies in fact that all such that must be in . We have that
which then implies for . Thus, we are able to make the aforementioned replacement, which then completes our proof.
Proposition 6 If is an open convex set containing , then its associated Minkowski functional is a subadditive function.
Proof: The two requirement of subadditive were proven in and respectively.
Theorem 1 Let be an open convex subset of a n.v.s . For arbitrary , there exists such that for all . In particular, the hyperplane separates and .
Proof: After a translation, we may always assume that . The Minkowski functional by Proposition 6 is a subadditive function. We define on to be linear. We can extend to on all of by the Hahn-Banach theorem (see [2] for details on it). Since by Proposition 4, for any , , we have for all , . Thus, setting gives us what we want to prove.
In [3], propositions of more general separation of sets by hyperplanes were proven. I shall not write them up here because I believe the foundational ideas behind their proofs have already been explained in the propositions above. I had learned of the Minkowski functional this week but initially did not feel like I grasped the motivation behind it. Such difficulty was resolved after reading 1.2 in [3], the title of which is “The Geometric Forms of the Hahn-Banach Theorem: Separation of Convex Sets”. Interestingly, I read about the Minkowski functional on Wikipedia before writing up [2], in the process of which I gained non-trivial understanding of the Hahn-Banach Theorem, which I had learned in 2017 or 2018 but forgotten in 2021 due to more or less superficial understanding. Certainly, the Hahn-Banach theorem can come across as very formal and abstract at first encounter, and it might not be immediate why it’s so significant. However, the geometric interpretation of it via separation of convex sets gives it more concrete meaning.
References
Originally published at 狗和留美者不得入内. You can comment here or there.
Theorem 1 (Baire Category Theorem) Let be a complete metric space.
Proof: Let be our intersection of a (countable) sequence of open dense sets. It suffices to prove that does not contain any nonempty open sets, or equivalently, that any open ball intersects with . By closure under finite intersection, we have that for all , is open. That is an open dense set implies that is open, which means it contains some closed ball of radius . With in mind, we also have that for some closed ball of radius less than . In this fashion we construct a sequence of closed balls for which , which implies that . Cantor’s intersection theorem (Theorem 1 of [2]) tells us that is non-empty, which means is non-empty, which completes our proof.
A set is nowhere dense iff its complement is dense in . Suppose by contradiction that is a collection of nowhere dense sets that such that . The application of de Morgan’s law tells us that , which the intersection of countably many dense open sets. This is impossible since is obviously not dense in .
Definition 1 If and are topological spaces, then a map is called open if is open in whenever is open in .
Proposition 1 For metric spaces and , is open iff for all any ball centered at , contains a ball centered at .
Proof: Suppose by contradiction that is not open yet any ball centered at is such that contains a ball centered at . In this case, there exists a ball centered at such that contains an element on its boundary, which we shall denote with . Every point has a open ball neighborhood contained in . By our hypothesis, must contain an open ball centered at that is also contained in , which is impossible since .
For the other direction, open implies that for any ball centered at , is open, which means it must contained a balled centered at .
Proposition 2 If and are normed vector spaces and is linear, then is open iff contains a ball centered at when is a ball of radius about .
Proof: A normed vector space is a metric space. Thus by Proposition 1, is open iff for all any ball over radius centered at , contains a ball centered at . We can openly and bijectively map to via . It is easy to see that
contains an open balled centered at iff is open.
Lemma 1 In a normed vector space , for any , .
Proof: Trivial from triangle inequality.
Lemma 2 Let be normed vector spaces. Let be a surjective linear map such that for all , contains a neighborhood of , or equivalently that every ball is contained by for some . If converges to and converges to , then .
Proof: By the argument in the proof of Proposition 2, we can replace in the hypothesis neighborhood of with neighborhood of and closed ball of radius centered at with the one centered at . Suppose by contradiction that . Because every neighborhood of has infinitely many points such that is arbitrarily close to , there exists some open ball centered at such that does not contain any neighborhood of . This open ball necessarily contains for any , . Thus, the hypothesis of the Lemma we are trying to prove has been violated.
Theorem 2 (Open mapping theorem) A surjective linear map between two Banach spaces is necessarily an open map.
Proof: Let denote the ball of radius centered at . By Proposition 2, it suffices to prove that for some , or equivalently, that for some , any such that implies that some satisfies .
That is surjective means that . By the Baire Category Theorem, some is not nowhere dense and thus its closure must contain an open ball. This tells us that must contain an open ball. Every non-zero corresponds to a one-dimensional subspace. The restriction of to that subspace is of course continuous. Thus, . For any , there is a neighborhood of such that . Thus, can not contain any such , which means that . This means that for some , .
We now attempt to show a slightly weaker statement than desired, namely that for some , . Any satisfies . That implies that has a larger diameter than . We already know that . We also observe that is equivalent to by the linearity of .
We want to upper bound a ball centered at . To do so, we note that is equivalent to
We also note that , which means by linearity of , . Thus, that implies that . This tells us that
Combining and gives us
which equates to
which tells us that suffices.
With Theorem 1 of [3] in mind, we try to construct a sequence such that for all and strict inequality for at least one , which means that with , converges to some such that . This is because if for any , we can construct such a sequence such that , our proof is complete. By linearity of , gives us for all ,
We notice that for any , there exists such that . Then, by , there exists such that . If ,
Let , the distance between and . If , then for such that is in . In this case, we can perturb and accordingly so that . Inductively, we derive that for any for all , there exists such that
Moreover, for each , the finite set of s associated with it contains the finite set of s associated with all such that . In this way, we have a sequence and a sequence of partial sums that converges to some . Via , the sequence , which converges to is induced. By Lemma 2, we have that . This completes our proof.
References
Originally published at 狗和留美者不得入内. You can comment here or there.
https://en.gravatar.com/gmachine1728
1729 = 10^3+7^3 = 12^3+1^3 is a number associated with Ramanujan. 1728 is also uniquely mathematical, in that it appears in the j-variant of the theory of complex multiplication. Though I have no actual detailed knowledge on the aforementioned mathematical topic as of June 2021, I shall quote the following.
Strong convergence in artistic tastes amongst people at the extreme right tail. Another point of convergence is the theory of complex multiplication in math, involving the j-invariant, moduli spaces, and abelian extensions of imaginary quadratic number fields.
For example, the great mathematician Barry Mazur writing
The elliptic modular function is loved by the analysts, arithmeticians and algebraic geometers who study elliptic curves since the isomorphism class of the elliptic curve formed by the lattice generated by the complex numbers 1 and z is completely determined by j(z), usually referred to as the j-invariant of the elliptic curve. It is the showcase example of a modulus in algebraic geometry, i.e., a continuous parameter that classifies a continuously varying array of distinct isomorphism classes of mathematical objects.
AND it was loved by Leopold Kronecker who hitched the aspirations of his youth (his Jugendtraum) on the ability of the elliptic modular function to help generate, in a magically ordered way, all algebraic numbers that are relatively abelian over quadratic imaginary number fields.
And David Hilbert saying
The theory of complex multiplication is not only the most beautiful part of mathematics but also of the whole of science.
And the great mathematician David Mumford writing:Especially, I became obsessed with a kind of passion flower in this garden, the moduli spaces of Riemann. I was always trying to find new angles from which I could see them better.
I wonder who this gmachine1728 is, and in case he sees this, he is welcome to use this contact page https://gmachine1729.wpcomstaging.com/%e8%81%94%e7%b3%bb-%d0%ba%d0%be%d0%bd%d1%82%d0%b0%d0%ba%d1%82-contact/.
Or email me at gmachine1729 at foxmail.com. I have had some interesting, high-quality people find me through this blog and contact me, and I also became good friends with some of them, chatting with them a fair bit on the internet. So yes, readers don’t be shy. Feel free to contact me if you have something worthwhile to talk about or something you want to ask. I like to think of myself as an easily approachable person. I will certainly respect your privacy and won’t leak your email information to anyone else without your permission.
Originally published at 狗和留美者不得入内. You can comment here or there.
The motivation behind the Hahn-Banach theorem can come across to a functional analysis newbie as somewhat elusive. I shall here try to explain this to the extent that I understand it.
Suppose that a seminorm on a vector space is such that iff , for some subspace of . Let be the dual space of . If is bounded with respect to this seminorm, then . We wish to induce via this seminorm a norm on the quotient space . Since a norm induces a metric (and a seminorm induces a pseudometric), it is natural then to define the norm on to be the distance corresponding to the seminorm between and , which is formally . It is easy to verify that this is well defined and a norm.
Similarly, induces an element of . Since is arbitrary, is also an arbitrary functional on its domain. We wish to show that given the constraint that for some subspace of , , we can for any , define functional on such that .
We are interested in extending a functional defined on a subspace to the full space with agreement of values on the subspace and a certain degree of boundnesses, more specifically an upper bound by the norm. Since this is trivially obtained by simply mapping to the elements outside the subspace, we are interested in an extension that is as non-zero or as large in absolute value as possible. In attempt to achieve this, we can try extending the functional with the requirement that its value on any input , which can be negative, is bounded above by the value of the application of another function on that has the reals, including negative ones, as its codomain. With this along with the properties of norms and seminorms in mind, we define the following.
Definition 1 Let be a real vector space. A sublinear functional on is a map such that for all and ,
We immediately notice that the constraints defining a sublinear functional are a subset of the constraints defining a seminorm or norm, which means that any seminorm or norm is necessarily a sublinear functional, which means that any proposition that holds for an arbitrary sublinear functional also holds for an arbitrary seminorm or norm.
Lemma 1 We extend a linear functional , where is a subspace of , defined such that for all , for an arbitrary sublinear functional to the subspace such that for all .
Proof: must of course also be nonzero. Let be our extended functional, with and . We require that for any , for any ,
The case of is trivial.
In the case of , the inequality in is equivalent to
In the case of , the inequality in is equivalent to
Here, we notice that the product of any by any scalar is also in , by the closure property of subspace. Thus, if we show that for arbitrary ,
we have shown the existence of the desired . follows from
in which we used linearity of , the fact that on , and the triangle inequality on . This completes our proof.
The proof of the Hahn-Banach theorem, of which Lemma 1, is the most difficulty part, can come across as coming out of the blue. I certainly developed a better idea of how to derive it by “working backwards”, as done above, first assuming the existence of the desired property, then finding a condition that implies it, and finally proving that that condition is indeed satisfied.
Theorem 1 (Hahn Banach theorem) Let be a real vector space on which is defined a sublinear functional . Let be any subspace of it and be some linear functional such that for all . Then, there exists a linear functional such that for all , and for all , .
Proof: Lemma 1 tells us that if , we can always extend onto some subspace , which is a proper extension of subspace such the extension of is bounded above by on and agrees with on . Let be the collection of two-tuples such that is a linear functional defined on and bounded above by on . Let . We say that iff and on . One easy verifies that this is a partial order on . For any chain in , we take the union of all sets in the chain and define a function with for any , for some in the chain that is defined on a domain that contains . It is apparent that for any in the chain, . Thus, we can apply Zorn’s lemma to derive the existence of a maximal element in with respect to this partial order. The set associated with any maximal element must be itself in order for Lemma 1 to not be violated.
Now, we will go about generalized the Hahn-Banach theorem to complex vector spaces.
Lemma 2 Let be a complex vector space and let be a linear functional on . If is a complex linear functional on and , then is a real linear functional, and for all . Conversely, if is a real linear functional on and is defined by , then is complex linear. In this case, if is normed, we have .
Proof: Let Then, for any , we write , where . We have and . Thus, . For any , . That for any , is also easily verified.
For the converse, one easily verifies that , and for ,
For any , we have that . This shows that . With for some , , since is linear, we have that . This shows that for any , there exists a of the same norm such that , which shows that . This completes our proof.
In the proof of the above lemma, we omitted the case of . [1] introduced the notation
Using this we can define the polar decomposition of any as
Applying to gives us . We note that in the proof of Lemma 2, we multiplied by .
Lemma 3 For any complex vector space , there exists a real vector space and a function that is bijective and linear with respect to real but not complex coefficients.
Proof: For any , we must have and also for all . We also stipulate that for any , .Take any basis of . Then we have as a set of basis elements defining , . To verify that no non-trivial linear combination of basis elements of can equal , one can simply use linearity to derive violation of the definition of basis of in the case of linear dependence of a subset of basis elements of .
We have defined to be linear with respect to real coefficients. It is not at all linear with respect to complex coefficients as and are basis elements of a real vector space; since the underlying field is multiplying a vector of it by an imaginary number is simply not defined here.
Theorem 2 (Complex Hahn-Banach Theorem) Let be a complex vector space, a seminorm on , a subspace of , and a complex linear functional on such that for . Then there exists a complex linear functional on such that for all and
Proof: Let . By Lemma 3, there exists a real vector space with linear with respect to real coefficients and bijective. Let be defined by . One easily verifies that is real linear functional on . Moreover, on , which is a subspace of , , with a seminorm on easily verified as well. By the Hahn-Banach theorem for real vector spaces (Theorem 1), there is a real linear functional defined on that agrees with on such that for all , . From this we also derive an analogous extension of to , which is . Now, let on . As in the proof of Lemma 2, if , we have . Since on , we also have on . This completes our proof.
Theorem 3 Let be a normed vector space (over ). Then,
Proof: We wish to define on , which is, by linearity, done simply by prescribing the value of . The function on is such that if and only if . We need to prescribe an and we try the largest value which satisfies the requirement as given in the Hahn-Banach theorem, where the take to be the upper bounding seminorm. We let . Take arbitrary . We now wish to show that
The inequality in this is equivalent to
which is true from the definition of .
We have by the definition of operator norm that
Set and pick a sequence such that from above which exists by definition of . We have thus shown that . Obviously, from this we also have . Applying Hahn-Banach theorem to with as the seminorm to the extension of shows that we can extend the extension of all of , which completes the proof of (1).
(2) is a special case of (1) with . As for (3), if , ether is a complex multiple of or not. If yes, would suffice. If not, we can define on to be the constant zero function, and then, by (1), we can extend to a function on the entire space such that .
References
Originally published at 狗和留美者不得入内. You can comment here or there.
Here, the underlying field of any vector space shall be either or . Moreover, subspace will always denote the subspace of a vector space.
Definition 1 A seminorm on a vector space over is a function that satisfies the following properties.
Proposition 1 For any seminorm , .
Proof: Follows directly from absolute homogeneity.
Definition 2 A norm on a vector space is a seminorm such that iff .
Definition 3 A vector space equipped with a norm is called a normed vector space. The topology it defines is called the norm topology on
Definition 4 A sequence of vectors in vector space converges with respect to norm iff .
Definition 5 A normed vector space that is complete with respect to the norm metric is called a Banach space.
Definition 6 A series converges absolutely iff .
Theorem 1 A normed vector space is complete iff every series in it that converges absolutely also converges with respect to the norm topology.
Proof: We assume the space is complete. This means that for any Cauchy sequence , for some . Now take any such that , which of course means that . To show that it converges, it suffices to show that is Cauchy. We have that for all
Originally published at 狗和留美者不得入内. You can comment here or there.
找到了一本俄罗斯人写的微分几何教材，当初看了觉得太枯燥，但近几天当弥补了不少自己
讲曲率之前，得先讲讲曲线的长度。
令 为某个曲线的闭弧， 为其参数化； 。我们注意到一个多边形线是 （ ）的一个由穿过某有序的有限的点集合 的相邻的点的线段构成的曲线。一个多边形线 是一个正则内接于曲线 的多边形当存在线段 的以点 的满足 的分割 。对每个多边形线对应其长度 。我们以 标记所有正则内接于曲线 的多边形线的集合。
定义 1.4.1 一个连续曲线 若 被称为可求长曲线。
定义 1.4.2 可求长曲线 的长度定义为 。
定理 1.4.1 光滑曲线的闭弧是可求长的，其长度为
证明：相当繁琐的用到区间分割的不等式估计。为了时间的考虑暂时不过。此证明也大概率类似
任意曲线若其所有闭弧都是可求长被称为可求长曲线。对可求长曲线，可以定义基于每一个闭弧的长度的存在的所谓的弧长参数化。取任意点 并联于 参数 的零值。为任意其他点 ，对应于等于的 的弧长 参数的值，若 在 之后我们给予其正符号 ，若 在 之前，我们给予其负符号 。若 有个光滑正则参数化 ，其弧长参数化也是光滑的，正则的。当考虑到符号，我们推导出弧长 。函数 是可导的， 。从而，存在反函数 ，其导数为
曲线 的弧长度（或单元速度）参数化定义于公式
从 ，我们得到矢量函数 的可导，其导数为
最后一个公式表明此弧长参数化的正则的。以弧长参数化 来表示，切矢量 ，主法矢量 和副法矢量 的形式简单如下：
第一个公式由 成立，第二个成立于等式
从这个，我们得到了 之间的正交。最后一个公式成立于矢量 的定义。
令 为 里的一个光滑的曲线。在此取一个点 ，另一个点 。我们以 标记在 的 的弧长，以 标记 在 和 切矢量 和。
定义 1.6.1 极限
若存在，叫做曲线 在点 的曲率。
我们会将曲线 在点 的曲率标记以 。
例子 1.6.1 (a) 若 是直线，在 的所有点， ， 。(b) 若 是半径为 的圆，很容易得到圆的所有点的曲率都是 。
定理 1.6.1 令 为 正则曲线。在其所有点都有曲率。若 是个 的正则参数化，则 .
证明：令 为 的弧长参数化，令 。从而， ， 为矢量 之间的角度。由于 ， ，故
以这些，我们证明了定理的断言曲率存在的第一部分，并得以公式
令 为任意 的正则参数化。利用 做个计算会得到
定理 1.6.2 在一个 正则曲线 的任意的点，密切平面的唯一存在的必要并且足够的条件是 的曲率在此点不等于零。
证明：从定理1.6.1，我们可以看到曲率不等于零当且仅当 之间不是平行的。在这种情况下，根据 [2] 里的 Theorem 2 的证明，只有一个密切平面存在。当曲率等于零时， ，故 及 ，一个直线。显然，任何包括直线的（多个）平面都是其密切平面。
定理 1.6.3 我们假设某个嵌入三维空间的平面包含所有曲线的点。令 为任意在线段 上的连续函数。曲率函数为 ， 为弧长参数的曲线 有唯一的存在（在由刚性运动而定义的等价关系下）。
证明：我们想寻求满足
的函数 。当我们不失去一般性地假设平面的法矢量为 ，从公式 ，我们能得到
然后不难发觉当
我们会得到
解 会给我们
其 为在初始点的切矢量 与 轴的角度。两个曲率一致的曲线的切矢量的变化也是一致的，则之间的切矢量的差是常矢量，位
习题 1.7.3 （平面曲线的Frenet公式）证明公式
同等于等式
解：一个简单的计算。
References
Originally published at 狗和留美者不得入内. You can comment here or there.
Here, we will be working in .
Proposition 1 The distance between a point and the plane given by is .
Proof: A normal vector of the plane is . We plug in to get
the solution of which is . Since every unit of corresponds to of distance, we have for our answer.
Proposition 2 The distance between a point and a straight line given by can be obtained by the magnitude of a cross product.
Proof: As for this distance, it is obtained by taking the perpendicular with respect the straight line that contains , which we shall call . We use to denote the distance between and . One notices that is equal to , where is the angle between the straight line given in the proposition and the straight line connecting and . We know that the magnitude of the cross product of two vectors is the product of their magnitudes and the sign of the angle between the two vectors, which completes our proof.
Definition 1 A regular curve is a connected subset of homeomorphic to some that is a line segment or a circle of radius . If the homeomorphism is in for and the rank of is maximal (equal to 1), then we say this curve is k-fold continuously differentiable. For , we say that is smooth.
Definition 2 Let a smooth curve be given by the parametric equations
The velocity vector of at is the derivative
The velocity vector field is the vector function . The speed of at is the length of the velocity vector.
Definition 3 The tangent line to a smooth curve at the point is the straight line through the point in the direction of the velocity vector .
We let denote the length of a chord of a curve joining the points and and denote the length of a perpendicular dropped from onto the tangent line to at the point .
Lemma 1 Let be continuous in . Then,
Proof: Trivial and left to the reader.
Theorem 1
Proof: We have that and by Proposition 2 that
We have, using properties of limits and keeping Lemma 1 in mind in the process,
Definition 4 A plane is called an osculating plane to a curve at a point if
Theorem 2 At each point of a regular curve of class where , there is an osculating plane , and the vectors are orthogonal to its unit normal vector .
Proof: Based on the following diagram from [1],
Originally published at 狗和留美者不得入内. You can comment here or there.
Theorem 1 (Chain rule) Let , , where and are open in , such that are differentiable on their respective domains. Then is also differentiable on , with for all .
Proof: We first assume that there exists a neighborhood of for which . This happens in the case of by inverse function theorem. In that case, by the definition of derivative and its properties, we have
In the case of , we have that for all ,
From this, we easily verifies that , which means that is differentiable at and in the case of , must hold as well.
Lemma 1 Let , be differentiable with and . Then,
Instead of , one can also use any closed interval of .
Proof: Follows directly from Fundamental Theorem of Calculus. See Theorem 2 (Newton-Leibniz axiom) of [1].
Lemma 1 is a statement of invariance of integral along parameterized smooth paths with the same endpoints.
Theorem 2 (Change of variables or u-substitution in integration) Let be any differentiable function of on , which is continuous on , and be Riemann integrable on intervals in its domain. Then,
Proof: Let be an antiderivative of . By the Fundamental Theorem of Calculus, it suffices to show that the left hand side of is equal to , which can be done by applying Lemma 1 accordingly.
Theorem 3 (Integration by parts) Let be differentiable functions on and continuous on . Then,
Proof: We have
Rearranging the above completes the proof.
References