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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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Proof that Frechet-metric generates same topology as the semi-norms

Given a countable family of semi-norms $p_i$, we can define a metric $d(f,g) = \sum \limits_{i=0}^{\infty} 2^{-i} \frac{ p_i(f-g) }{ 1 + p_i(f-g) }$ We have the locally convex topology induced by the semi-norms as above, as well as the topology…
shuhalo
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multiplicative semi-norms on $\mathbb{C}[x]$

A multiplicative semi-norm on a ring $A$ is a function $|\,|:A\to \mathbb{R}_{\ge 0}$ that is multiplicative and satisfies the semi-norm conditions: $|0|=0,|1|=1\\ |fg|=|f||g|,\\ |f+g|\le |f|+|g|.$ I want to see why the set of multiplicative…
Matt
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Applications of the Closed graph theorem

In functional analysis a famous theorem states that: if $X, Y$ are Banach spaces and $T: X \to Y$ is a linear operator, $T$ is continous if and only if the graph $\Gamma_T:={(x,Tx), x \in X}$ is closed in the product topology. Do you know some nice…
user137921
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Proofs that rely on an infinite matrix

If I have an operator $A\in B(\mathcal{H})$ that can be "identified" with an infinite matrix with countably many entries, is it in any way unrigorous to do actual calculations with the picture we have in mind of finite matrices. i.e. is it wrong to…
Squirtle
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Existence of functional implies non-negative supremum.

$\ell_\infty(T) = \{f:T\to\mathbb{R} : \sup_{t \in T}|f(t)| < \infty\}$. Let $G$ be a family of transformation of set $T$ and $V$ linear subspace of $\ell_\infty(T)$ such that: $1_T \in V$, $\forall f, g \in V \Rightarrow f\circ g \in V$. Show that…
Stephen Dedalus
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T is not compact and orthonormal sequence

I want to show that if $\,T$ is not compact then there exists an orthonormal sequence $x_{n}$ and $R>0$ such that $ \forall n\in \mathbb{N}\,\,\,\,\|T(x_{n})\|\geq R$. It is obvious by the definition of compact operator that we can find a sequence…
user185559
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bounded operator $T$ is not compact then there exists an orthonormal sequence $e_n$ and $d>0$ such that $\|T(e_n)\|>d$ for all $n\in\Bbb{N}$?

I want to prove that if a bounded operator $T$ is not compact then there exists an orthonormal sequence $e_n$ and $d>0$ such that $\|T(e_n)\|>d$ for all $n\in\Bbb{N}$. Could someone helps me? I think that the fact that T is not compact implies that…
user185559
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Nonexistence of a certain norm on $C[0,1]$

Question: Let $X=C[0,1]$, show that there is no such norm $\lVert\cdot\rVert_*$ on $X$ that for any series $\{f_n\}_{n=1}^{\infty}\subset X$, $$\lim_{n\to\infty}\lVert f_n\rVert_*\to 0\Longleftrightarrow \lim_{n\to\infty}f_n(t)=0,\quad\forall…
NGY
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Is c, the Banach space of convergent sequences with the sup norm, separable?

Is $c$, the Banach space of convergent sequences with the sup norm, separable? Let $X$ be the set of all sequences which are rational numbers, that converge to some rational number $x$. As the rationals are countably infinite, we need to only show…
user3784030
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Find a space whose dual does not separate points

I read about the fact that for a locally convex topological vector space $X$, its dual $X^*$ separates points, i.e. for any $x\neq y$ in $X$, $\exists f \in X^*$ such that $f(x)\neq f(y)$. Could you help me to find a non locally convex topological…
Petite Etincelle
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Show that if $X$ is a Hilbert space, then so is $Y$

Given that $X$ and $Y$ are normed linear spaces and that $T:X\to Y$ is a linear map, such that $T(\alpha x_1 + \beta x_2) = \alpha T(x_1)+ \beta T(x_2)$ for all vectors $x_1,x_2\in X$ and scalars $\alpha , \beta$. Suppose $T$ maps $X$ onto $Y$ and…
user3784030
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Stone's theorem

I have some basic doubts about Stone's theorem. 1) Can we apply Stone's theorem to conclude that given any Unitary operator U, we can find a self adjoint operator A such that U = exp(i A). That is, is any unitary operator part of a one parameter…
user157106
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$C([0,1])$ is not weakly sequentially complete.

I'm studying Functional Analysis by myself. For a counterexample of every Banach space is not weakly sequentially complete, I was suggested to check $C([0,1])$ is not weakly sequentially complete. For this, let $f_n(t)=1-nt$ if $0\leq t\leq…
niki
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Proof of a theorem about Baire categories

Problem: prove that the set of $C([0, 1])$ functions whose derivative is defined at every point (and it is either finite or infinite) is of the first Baire category. I have no idea how to approach this and would be very grateful for any help.
Demons94
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Inverse of identity operator not continuous

Let $\mathbb 1: (C^1([0,1]), \|f\|:=\|f\|_\infty + \|f'\|_\infty)\to (C^1([0,1]),\|\cdot\|_\infty)$ denote the identity mapping between $C^1([0,1])$ with different norms. Then $f$ is linear, continuous and one-to-one, but the inverse Operator…
dinosaur
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