The special planar version of Divergence Theorem is actually equivalent to Green’s Theorem, even though on first sight they look quite different.
Abel’s Theorem is a useful theorem in analysis.
Let be any sequence in or . Let . Suppose that the series converges. Then
where is real, or more generally, lies within any Stolz angle, i.e.\ a region of the open unit disk where for some .
One quick way is to use Caratheodory’s Criterion:
Let denote the Lebesgue outer measure on , and let . Then is Lebesgue measurable if and only if for every .
Suppose is a set with outer measure zero, and be any subset of .
Then by the monotonicity of outer measure.
The other direction follows by countable subadditivity of outer measure.
This is a list of miscellaneous theorems from Functional Analysis.
Hahn-Banach: Let be a real-valued function on a normed linear space with
(i) Positive homogeneity
Let denote a linear subspace of on which for all . Then can be extended to all of : for all .
Geometric Hahn-Banach / Hyperplane Separation Theorem: Let be a nonempty convex subset, . Let be a point outside . Then there exists a linear functional (depending on ) such that for all ; .
Riesz Representation Theorem: Let be a linear functional on a Hilbert space that is bounded: . Then for some unique .
Principle of Uniform Boundedness: A weakly convergent sequence is uniformly bounded in norm.
Open Mapping Principle/Theorem: Let be Banach spaces. Let be a bounded linear map onto all of . Then maps open sets onto open sets.
Closed Graph Theorem: Let be Banach spaces, be a closed linear map. Then is continuous.
Theorem 11 (Lax Functional Analysis): A weak* convergent sequence of points in a Banach space is uniformly bounded.
We will need a previous Theorem 3: is a Banach space, a collection of bounded linear functionals such that at every point of , for all . Then there is a constant such that for all .
Sketch of proof:
Weak* convergence means , thus there exists such that for all , we have , which in turns means via the triangle inequality. We have managed to bound the terms greater than equals to .
For those terms less than , we have .
Thus, we may take . The crucial thing is that depends only on , not .
Then, use Theorem 3, we can conclude that for all .
The sequence converges weakly to means that for every linear functional in .
The notation for weak convergence is , which is typed as \rightharpoonup in LaTeX.
Hahn-Banach Theorem is called the Crown Jewel of Functional Analysis, and has many different versions.
There is a Chinese quote “实变函数学十遍，泛函分析心犯寒”, which means one needs to study real function theory ten times before understanding, and the heart can go cold when studying functional analysis, which shows how deep is this subject.
The following is one version of Hahn-Banach Theorem that I find quite useful:
(Hahn-Banach, Version) If is a normal vector space with linear subspace (not necessarily closed) and if is an element of not in the closure of , then there exists a continuous linear map with for all , , and .
Brief sketch of proof: Define , , and use the Hahn-Banach (Extension version).
Let be normed linear space, a subspace of . The closure of , , is a linear subspace of .
We use the “sequential” equivalent definition of closure, rather than the one using open balls: is the set of all limits of all convergent sequences of points in . Let , . There is a sequence in such that . Similarly there is a sequence in which converges to .
Then is a sequence in that converges to . is a sequence in that converges to .
If we have two normed linear spaces and , their Cartesian product can also be normed, such as by setting , , or . Note that we are following Lax’s Functional Analysis, where a norm is denoted as , rather than which is clearer but more cumbersome to write.
It is routine to check that all the above 3 are norms, satisfying the positivity, subadditivity, and homogeneity axioms. Minkowski’s inequality is useful to prove the subadditivity of the last norm.
We may check that all of the above 3 norms are equivalent. This follows from the inequalities , and , where . In general, we have that all norms are equivalent in finite dimensional spaces.
This theorem can be considered a converse of a previous theorem.
Theorem: Let denote a positive homogenous, subadditive function defined on a linear space over the reals.
(i) The set of points satisfying is a convex subset of , and 0 is an interior point of it.
(ii) The set of points satisfying is a convex subset of .
Proof: (i) Let . Let . For ,
Therefore is convex. We also have .
The proof of (ii) is similar.
The Arzela-Ascoli Theorem is a rather formidable-sounding theorem that gives a necessary and sufficient condition for a sequence of real-valued continuous functions on a closed and bounded interval to have a uniformly convergent subsequence.
Statement: Let be a uniformly bounded and equicontinuous sequence of real-valued continuous functions defined on a closed and bounded interval . Then there exists a subsequence that converges uniformly.
The converse of the Arzela-Ascoli Theorem is also true, in the sense that if every subsequence of has a uniformly convergent subsequence, then is uniformly bounded and equicontinuous.
Explanation of terms used: A sequence of functions on is uniformly bounded if there is a number such that for all and all . The sequence is equicontinous if, for all , there exists such that whenever for all functions in the sequence. The key point here is that a single (depending solely on ) works for the entire family of functions.
Let be a continuous function and let be a sequence of functions such that
Prove that there exists a continuous function such that for all .
The idea is to use Arzela-Ascoli Theorem. Hence, we need to show that is uniformly bounded and equicontinuous.
This shows that the sequence is uniformly bounded.
Similarly if , .
If and ,
Therefore we may choose , then whenever , . Thus the sequence is indeed equicontinuous.
By Arzela-Ascoli Theorem, there exists a subsequence that is uniformly convergent.
By the Uniform Limit Theorem, is continuous since each is continuous.
This is a basic example of a function of bounded variation on [0,1] but not continuous on [0,1].
The key Theorem regarding functions of bounded variation is Jordan’s Theorem: A function is of bounded variation on the closed bounded interval [a,b] iff it is the difference of two increasing functions on [a,b].
Both and are increasing functions on [0,1]. Thus by Jordan’s Theorem, is a function of bounded variation, but it is certainly not continuous on [0,1]!
Question: Let belong to both and , with . Show that for all .
There is a pretty neat trick to do this question, known as the “interpolation technique”. The proof is as follows.
For , there exists such that . This is the key “interpolation step”. Once we have this, everything flows smoothly with the help of Holder’s inequality.
Note that the magical thing about the interpolation technique is that and are Holder conjugates, since is easily verified.
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Holder’s Inequality is a very useful inequality in Functional Analysis, hence many results can be proved by applying Holder’s Inequality.
Suppose that and . Prove that if in , then in .
Proof: Assume in . Then there exists such that if , then .
Since is arbitrary, as .