PEOPLE WHO KNOW FUNCTIONAL ANALYSIS, GET IN HERE NOW

>>8519111
Just read the damn book and then kys

>>8519111
The only thing I know in functionnal analysis are 2 ways to prove a Fourrier series is convergeant, how to deduce a function from its Fourrier series, and how to form a Fourrier series from a periodic function

Should I kms ?

>>8519262
Sounds like you mean "function analysis" or something, because what you just described is barely related.

>>8519265
Probably
What is the difference ?

>>8519270
Functionals are specific kinds of functions. Functions that map functions to numbers.

>>8519270
Well, one provides an alternative way to deal with functions and the other does everything described in OP, none of which explicitly involve functions. The closest is L^2 I guess since you'd usually call them L^2-functions if I remember correctly.

>>8519273
>>8519275
Thanks. So yeah, I know nothing about them.

>>8519265
Functional analysis isn't really the analysis of functionals, nowadays it refers to the analysis of topological vector spaces, and that includes Fourier analysis, Banach algebras, operator theory, etc...
It started as the analysis of the spaces of functions and is related to the development of the calculus of variations, and I think that's explains the word functional.

All right, I'll try to frantically remember my functional analysis:
> Baire category, open mapping thm, the uniform boundedness principle
Baire: In a countable metric space, every countable intersection of dense open sets is dense.
Proof: Pick a countable family of open sets [math](X_n)[/math] and a nonempty open set [math]U[/math]. There is a [math]x_1 \in U \cap X_1[/math] and [math]r_1 < 1[/math] such that [math]\bar B(x_1, r_1) \subset U \cap X_1[/math]. Now, there is an [math]x_2 \in B(x_1, r_1) \cap X_2[/math] and [math]r_2 < 1/2[/math] such that [math]\bar B(x_2, r_2) \subset B(x_1, r_1) \cap X_2[/math].
Continuing like this, we get a sequence [math](x_n)[/math] such that [math]d(x_n, x_{n+1}) < 1/2^n[/math] and [math]x_n \in X_1 \cap \dots X_n[/math].
This sequence is Cauchy and converges to a point [math]x[/math] that is in [math]\bar B(x_n, r_n)[/math] for each n, hence in [math]U\cap \bigcap X_n[/math]

Open mapping: Every surjective linear map [math]f: E \to F[/math] between banach spaces is open.
Proof: We need only prove that the image of each open ball contains an open ball. Now, since translations are homeomorphisms, we need only prove that the image of each open ball centered at 0 contains an open ball.
Let [math]r > 0[/math]. Since f is surjective, we can write [math]\bigcup_n n\overline{f(B_E(0, r))} = \bigcup \overline{f(B_E(0,nr))} = F[/math].
Baire's theorem tells us that one of these closed sets has nonempty interior, hence [math]\overline{f(B_E(0,r))}[/math] has nonempty interior. Better, we can find [math]s(r) > 0[/math] st [math]B_F(0, s(r)) \subset \overline{f(B_E(0,r))}[/math] (use the fact that [math]\overline{f(B_E(0,r/2))} - \overline{f(B_E(0,r/2)) \subset \overline{f(B_E(0,r))[/math]).
Now we only need to remove those bars, but honestly I don't remember how.

>>8519922
The thing to remember about Baire is that it allows you to perform quantifier inversion, case in point:

uniform boundedness principle: Let [math]S[/math] be a set of linear maps between Banach spaces E and F. Then either [math]\sup_{\phi \in S}|||\phi||| < \infty[/math] or there is an [math]x \in E[/math] such that [math]\sup_{\phi \in S}||\phi(x)||_F = \infty[/math].

This tells you that if you have pointwise boundedness (for each x, there exists a bound on the phi(x)), you have uniform boundedness (there exists a bound M such that for each x, |phi(x)| < M)

Proof: Assume [math]\sup_{\phi \in S} ||\phi(x)||_F < \infty[/math] for each x in E. Consider the set [math]K = \{x \in E | \foreach \phi \in S ||phi(x)||_F \le 1\}[/math]. This is a closed subset of E and, since [math]nK = \{x \in E | \foreach \phi \in S ||phi(x)||_F \le n\}[/math], we can write [math]E = \bigcup_n nK[/math].
Using Baire's theorem, we now know that K has nonempty interior. Hence, there are [math]x \in E, r > 0[/math] such that for each [math]u \in B(0,1)[/math] [math]||\phi(x + ru)||_F \le 1[/math] for each phi, and therefore [math]\sup_{\phi \in S} |||\phi||| \le \frac{1}{r}(\phi(x) + 1) \infty[/math].

Baire creates breaches, that you have to exploit

Any specific things you want to know? I'm taking graduate functional analysis now and have pretty close to a perfect score in the class. It's basically just linear algebra in infinite dimensions, which sort of surprisingly somehow doesn't make things all that much less well-behaved. Also, FYI for people getting confused about the name, functionals are functions from a vector space to its underlying field. In functional analysis, we basically only care about linear functionals, which express a nice duality to the original vector space structure. This mostly comes to play via the fact that spaces with virtually no structure have duals with quite a bit of structure -- namely in the sense that almost any space you might care about has a dual which is a Banach space (i.e. a vector space with a norm which is complete with respect to that norm).

>tfw too dumb to learn FA
is Kreyszigs book good? or is it watered down?

>>8519977

> Hahn Banach Extension
Let V be a real vector space and [math]f: V \to \mathbb R[/math] such that [math]f(tv) = tf(v)[/math] for each positive t and [math]f(v + w) \le f(v) + f(w)[/math].
If [math]\phi[/math] is a linear form on a subspace [math]U\subset V[/math] and [math]\phi \le f_{|U}[/math], then there is a linear extension [math]\psi: V \to \mathbb R[/math] that still satisfies [math]\psi \le f[/math].

The main difficulty resides in the finite dimensional case (how to extend from dimension n to n+1, ie. what is done at the end), the infinite-dimensional one being a purely formal application of Zorn's lemma

Proof: Consider the set [math]\{(W, \psi)| U \subset W \subset V, \psi_{|U} = \phi\}[/math] ordered by [math](W, \psi) \le (X, \theta)[/math] iff [math]W \subset X[/math] and [math]\theta_{|W} = \psi[/math].
This is an inductive set and it has a maximal element [math](W, \psi)[/math].
Assume [math] W \ne V[/math]. Then we can find [math]v \in V \setminus W[/math]. Now, for each [math]x,y\in V[/math], we can write [math]\psi(x+y) \le f(x-v) + f(y+v)[/math] and thus [math]\psi(x) - f(x-v) \le f(y+v) - \psi(y)[/math]. Picking [math]\alpha[/math] such that [math]\sup_{x\in W} \psi(x) - f(x-v) \le \alpha \le \inf_{y \in W} f(y+v) - \psi(y)[/math], we can set [math]\tilde \psi[/math] on [math]W \oplus \mathbb Rv[/math] by [math]\tilde \psi(w + tv) = \psi(w) + t\alpha[/math] and check that we now have [math](W,\psi) < (W \oplus \mathbb Rv, \tilde \psi)[/math], which contradicts the maximality. Hence V=W.

It has a number of corollaries, for example:
. The weak topology on a locally convex space is Hausdorff
.If E is a normed linear space, then for each x, [math]||x|| = \sup_{\phi \in B_{E'}(0,1)} |\phi(x)|[/math]
. A subspace U of a lctvs V is dense iff for each functional [math]\phi \in V'[/math], [math]\phi_{|U} = 0 \Rightarrow \phi = 0[/math]

I just got done with a functional analysis course and honestly I don't know how to use anything I learned in it.

Example, I know the Hahn Banach Theorem and its basic results but I honestly don't know why it matters.

Basic Real Analysis/Measure Theory is obviously useful, functional analysis I'm not so sure about.

>>8520051
basically, PDE is a huge application of functional analysis, i.e. find the right function space with the right norm, state your equation as Af =g with A some differential operator and try to solve it using Anal Func

>>8520080

That makes sense, pde aren't really a big interest of mine though.

Also seems it's restricted to linear PDE

>>8520051
In math, the first subjects that come to mind are PDE and Noncommutative Geometry.

In physics, anything involving Quantum Mechanics.

>>8520101
Also dynamical systems.

>>8520033
> Duals of Hilbert spaces
First step in the Riesz representation theorem is the following remark:
If C is a closed convex subset of a Hilbert space H, then for each x in H, there exists a unique p(x) such that [math]||x-p(x)|| = d(x, C)[/math]. It is also characterized by the property [math]\langle p(x)-x, p(x)-y \rangle \le 0[/math] for each y in C.
Proof: Consider a sequence [math](x_n) \in C^{\mathbb N}[/math] such that [math]||x-x_n|| \to d(x, C)[/math]. We'll prove that this sequence is Cauchy.
The idea is to use the parallelogram identity: [math]||x_p+x_q- 2x||^2 + ||x_p-x_q||^2 = 2(||x_p- x||^2 + ||x_q-x||^2)[/math] and then write [math]||x_p - x_q||^2 = 2(||x-x_p||^2 + ||x-x_q||^2) - 4\left|\left| x-\frac{x_p+x_q}{2}\right|\right|^2[/math]. Since C is convex, we can write [math]||x_p - x_q||^2 \le 2(||x-x_p||^2 + ||x-x_q||^2) - 4d(x,C)^2[/math].
Let [math]\epsilon > 0[/math]. There is a [math]N \in \mathbb N[/math] such that [math] p > N \rightarrow ||x_p-x|| \le d(x,C) + \epsilon[/math].
Now, for p,q >N, [math]||x_p - x_q||^2 \le 8d(x,C)\epsilon + 4\epsilon^2[/math], hence [math]x_p[/math] is Cauchy and converges to a point [math] p(x) \in C[/math] (because C is closed).
Now, if y is in C, we write [math]||x-p(x)||^2 \le ||x - (tp(x) + (1-t)y||^2[/math] for each t in [0,1], expand and deduce [math]\langle x-p(x), y-p(x) \rangle \le 0[/math].
It's then easy to see that this implies that p(x) is uniquely determined.

In the case when C is a closed subspace of H, p is actually an orthogonal projection operator, which proves the following important theorem:
If [math]V[/math] is a closed subspace of [math]H[/math], then we have [math]H = V \oplus V^{\perp}[/math] and [math]V^{\perp}[/math] is a also closed subspace.

>>8520114

It has as a corollary Riesz's theorem: The map [math]\Phi:\begin{array}{c c c} H & \to & H' \\ a & \mapsto & \left\{ \begin{array}{c c c } H & \to & \mathbb R \\ x & \mapsto & \langle a,x \rangle \end{array} \right\} \end{array}[/math] is an isometric isomorphism.
Indeed, this map is clearly an isometry. Now, given a form [math]\phi \in H'\setminus\{0\}[/math], [math](\ker \phi)^{\perp}[/math] is nonzero (because [math]\ker \phi[/math] is closed) and hence generated by a vector [math] a \in H \setminus\{0\}[/math]. Now, the forms [math]\phi[/math] and [math]\Phi(a)[/math] have the same kernel, hence there is a [math]\lambda[/math] such that[math]\phi = \lambda \Phi(a) = \Phi(\lambda a)[/math], qed.

>>8520094
>Also seems it's restricted to linear PDE
It's not, it's just that you probably won't get to nonlinear aspects of functional analysis in your first course.

>>8520051
functional analysis is useful for PDE, operator algebras, probability, geometric group theory, pretty much anything that has to deal with convexity or measure theory really

>>8519991
> It's basically just linear algebra in infinite dimensions, which sort of surprisingly somehow doesn't make things all that much less well-behaved.

Want to know how I can tell your class isn't very comprehensive?