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Questions tagged [functional-analysis]

Functional analysis, the study of infinite-dimensional vector spaces, often with additional structures (inner product, norm, topology), with typical examples given by function spaces. The subject also includes the study of linear and non-linear operators on these spaces and other topics. For basic questions about functions use more suitable tags like (functions), (functional-equations) or (elementary-set-theory).

Functional analysis is the study of infinite-dimensional vector spaces, often with additional structures (inner product, norm, topology), with typical examples given by function spaces. The subject also includes the study of linear and non-linear operators on these spaces, including spectral theory, as well as measure, integration, probability on infinite dimensions, and also manifolds with local structure modeled by these vector spaces.

For basic questions about functions use more suitable tags like , or .

52582 questions
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Distinguishing between symmetric, Hermitian and self-adjoint operators

I am permanently confused about the distinction between Hermitian and self-adjoint operators in an infinite-dimensional space. The preceding statement may even be ill-defined. My confusion is due to consulting Wikipedia, upon which action I have the…
Josef K.
  • 799
31
votes
2 answers

Showing that $\ker T$ is closed if and only if $T$ is continuous.

Possible Duplicate: $T$ is continuous if and only if $\ker T$ is closed Let $T: X\to \mathbf{R}$ be linear. Suppose that $X$ is a Banach space. I want to show that $T$ is continuous if and only if $\ker T $ is closed. My Attempt.…
Nana
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30
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1 answer

$C(X)$ is separable when $X$ is compact?

$X$ is a compact metric space, then $C(X)$ is separable, where $C(X)$ denotes the space of continuous functions on $X$. How to prove it? And if $X$ is just a compact Hausdorff space, then is $C(X)$ still separable? Or if $X$ is just a compact (not…
David Chan
  • 1,960
29
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1 answer

weak sequential continuity of linear operators

Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this imply that…
user1736
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28
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2 answers

Every weakly convergent sequence is bounded

Theorem: Every weakly convergent sequence in X is bounded. Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$,…
luka5z
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28
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5 answers

Does there exist a linearly independent and dense subset?

Do there exist in infinitely dimensional normed spaces linearly independent and dense subsets? (Existence of linearly independent dense subset is equivalent of existence of dense Hamel Basis.) Thanks.
Richard
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28
votes
3 answers

Proving that closed (and open) balls are convex

Let $X$ be a normed linear space, $x\in X$ and $r>0$. Define the open and closed ball centered at $x$ as $$ B(x, r) = \{y \in X : \Vert x − y\Vert < r\} $$ $$ \overline{B}(x, r) = \{y \in X : \Vert x − y\Vert \leq r\}. $$ Then $B(x, r)$ and…
Adam Rubinson
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24
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2 answers

Is there a chain rule for functional derivatives?

Given a functional $S=S\{Y[X(r)]\}$, is the following "chain rule" valid? $$\frac{\delta S\{Y[X]\}}{\delta X(r)}=\frac{\partial Y(r)}{\partial X(r)}\frac{\delta S[Y]}{\delta Y(r)}$$
Machine
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23
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2 answers

Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?

Let $X$ and $Y$ be Banach spaces. A bounded operator $T\colon X\to Y$ is called Fredholm iff The dimension of $\ker(T)$ is finite, The codimension of the image $\mathrm{im}(T)$ is finite, The image $\mathrm{im}(T)$ is closed in $Y$. Question: Is…
Rasmus
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21
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4 answers

Why do we need the Hahn-Banach Theorem to extend a bounded linear functional?

I'm beginning with functional analysis and I have a related question. It concerns the Hahn-Banach theorem. In particular, I cannot appreciate its value, most probably because I don't understand it. I mean the Hahn-Banach theorem in it's most common…
gpo
  • 749
20
votes
4 answers

Nonlinear function continuous but not bounded

I would like an example of a map $f:H\rightarrow R$, where $H$ is a (infinite dimensional) Hilbert space, and $R$ is the real numbers, such that $f$ is continuous, but $f$ is not bounded on the close unit ball $\{ x\in H : \|x\| \leq 1\}$. Actually,…
Matthew Daws
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19
votes
1 answer

Hilbert spaces and unique extensions of linear functions.

So I'm pondering the statement: Show that a continuous linear functional $f$ on a subspace $V$ of a Hilbert space $H$ has a unique norm preserving extension $h$ on $H$ Here is my thought. Consider the closure of $V$, $\overline{V}$. Then since $V$…
user58514
17
votes
1 answer

Intuition for the Lax-Milgram Theorem

I have seen the proof of the Lax-Milgram theorem (and can replicate it with most of the details), I've also applied it for a bi-linear form defined as the weak formulation of a PDE to show the existence of a solution. I still don't get the idea…
user311475
17
votes
1 answer

Mazur's Lemma (can't find a proof anywhere)

If we have a normed vector space $X$, and $x_n \to x$ weakly, then it's clear how to show there's a sequence $y_n \to x$ strongly (since the weak closure and strong closure of a convex set coincide, just consider the closure of the convex hull of…
JohnathanChen
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16
votes
2 answers

Intuition on an unexpected finite dimensional space

We had this exercise that uses Riesz' lemma to prove that a functional space actually has a finite dimension. I was curious to know what this space "looks" like, if we can find natural elements that would generate it. Here are the assumptions: $E =…
Barnabé Monnot
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