New answers tagged functional-analysis
1
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Accepted
Proof of the completion of a normed space
Recall that elements of $\hat{X}$ are equivalence classes of Cauchy sequences in $X$, with the equivalence relation $[x] \sim [y]$ iff $x_n - y_n \to 0$, i.e $\lim_{n \to \infty}\|x_n - y_n\| = 0$. ...
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Do we have Arzela-Ascoli theorem on the space of all continuous stochastic process?
$X_k$ need not have a subsequence that converges in $U$. One way to generate a counterexample is to consider constant processes (i.e. processes such that $t \mapsto X_{k,t}(\omega)$ is constant for ...
- 19k
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Semigroup properties of spectral fractional Laplacian
Yes on all counts. The first question should be whether the spectral fractional Laplacian generates a semigroup at all. This is indeed the case. An explicit formula is
$$
P_t f=\sum_{k=1}^\infty e^{-t\...
- 14.7k
1
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Accepted
When you extend a bounded, injective, and compact operator with dense image, is the extension also injective
Let $X = Y = \ell^2$, $c_c$ are the finite sequences and
$$
V = \left\{x \in c_c \;\middle|\; x_1 = \sum_{n=2}^\infty x_n \right\}.$$
Further,
$$
L x := (x_2/2, x_3/3, x_4/4, \ldots).$$
It is clear ...
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If $(f_{n})_{n \in \mathbb{N}}$ is bounded in $\mathcal{C}([0,1],\mathbb{R})$, then $(Sf_{n})_{n \in \mathbb{N}}$ has a convergent subsequence
The result has to be true, because $S$ (the Volterra Operator) is compact. Here is an ad-hoc argument.
For each $n\in\mathbb N$, consider the partition $X_n=\big\{[\frac{k-1}n,\frac kn)\big\}_{k=1}^n$ ...
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-2
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Sharp constant in the $L^p$ regularity estimate?
That is just an attempt, I tried using Fourier theory and the Hardy–Littlewood–Sobolev theorem, but this does not work if $\alpha=2$ is equal to the dimension $d=2$, where $\alpha$ is the power ...
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Is there an inequality to compare the essential supremum of a function to the function itself?
Yes, it follows that $\text{ess sup} f \geq f(x)$ for almost every $x\in \Omega$.
Indeed, if by contradiction this is false, then there exists a set $A$ of positive measure so that $f(x) > \text{...
- 13.3k
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How can I show $\|XMY\|=\|M\|$?
The only way I can prove it is using tensor notation.
Start with the definition,
$$
\|M\|^2 = {M^j}_i {{M^*}_j}^i = {M_j}^i {{M^*}^j}_i
$$
since if $M = {M^j}_i$ then $M^H := {{M^*}_j}^i$ (in words, a ...
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Accepted
How can I show $\|XMY\|=\|M\|$?
I'm afraid your second step $\langle XMY, XMY\rangle=\langle X^HXM,MYY^H\rangle $ is not valid, since we can only do the conversion $X\to X^H$ from the left side, not the right side. The correct proof ...
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Exponential Shift Operator with another differential Operator
If you Fourier transform by $y$, you will get $$e^{{1\over 2} y^TCy}\sim \det(C)^{-d\over2}\int e^{iqy}e^{{1\over2}q^{T}C^{-1}q}dq$$ and so $$e^{-t\nabla_x \nabla_y} e^{{1\over2} y^TCy}\sim\int dq e^{...
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Commutant of a set of operators and norm topology.
Note the weak topology is weaker than the strong topology which in turn is weaker than the norm topology. So saying the commutant is WOT closed implies it is closed in the other two (this is in ...
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Hankel Operator is compact
Solution 1 By the Schur test, with $p_n=1,$ we have $\|A\|_{\ell^2\to \ell^2}\le \sum_{n=1}^\infty|a_n|.$ Let $A_N$ denote the truncated matrix, i.e.
$$A_N(i,j)=\begin{cases} A(i,j) & i+j\le N\\
0 ...
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2
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Using density to prove an inequality for functions $f \in H^{1}(\mathbb{R}^{d})$
First: Fatou’s lemma is the key. Whenever you have convergence in $L^2$, you have convergence a.e. (up to subsequences). And whenever you have a sequence of functions $\varphi_n$ that converge a.e. to ...
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0
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Ultra-weak topology on space of operators: conditions for convex set to be closed
Ok, I guess to close the question I should just copy down the guided exercise in Rudin's 'Functional Analysis' book, chapter 4, exercise 21.
Here $X$ is a Banach space with its dual of continuous ...
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Maximal domain of unbounded linear differential operator
To determine the domain $\mathcal{D}(\mathrm{T})$ and the range $\mathcal{R}(\mathrm{T})$ of the linear operator $\mathrm{T}$ defined as $\mathrm{T}f = v \cdot \partial_xf(x,v)$ on the weighted ...
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Are nuclear operator between Banach spaces compact?
The $T_{m}$ that you have defined are finite rank operators, which are compact since they have finite dimensional range. Since you have showed that the the operator $T$ is a norm limit of finite rank ...
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Counterexample concerning Rudin's definition of open sets in $\mathscr D(\Omega)$
The simplest counterexample $\{0\}$ was given in comments.
Since you asked meanwhile for "a convex balanced set $W⊂\mathcal D$
such that $\mathcal D_K∩W$
is open for some $K,$ but not all $K$&...
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0
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Space of functions of infinitely many variables with norm different from the uniform one
Assuming each $\mu_t$ is a Radon probability (your seminorm $\|\cdot\|_p$ is ill-defined without some assumption on the total masses of the measures $\mu_t$), one can prove the existence and ...
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Is the derivative of differentiable function $f:\mathbb{R}\to\mathbb{R}$ measurable on $\mathbb{R}$?
I have read the only answer posted to this question, but I am afraid it is not right:
Of course, if the derivative of $f$ at a point $x$ is greater than a real number $L$, there will be a natural ...
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4
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Cauchy-Schwarz inequality for Bochner spaces
Holder’s inequality very generally states that for every measurable function $f,g:\Omega\to [0,\infty]$, we have for every $1\leq p\leq\infty$ that
\begin{align}
\int_{\Omega}fg\,d\mu\leq \|f\|_{p}\|g\...
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Counterexample concerning Rudin's definition of open sets in $\mathscr D(\Omega)$
As noted by Anne in the comments, there are several easy counterexamples ($\{0\}$ being one of them). However, I think you are a little confused when it comes to the definition here.
The definition ...
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Is there a proof that continuous linear operators are bounded that uses this line of reasoning?
Firstly, unfortunately there is no such result about continuous images of unit spheres being bounded in infinite dimensions. In fact, a result of Bessaga shows that an infinite dimensional Hilbert ...
- 4,002
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Holder inequality for 3 functions
Indeed, you can recursively apply the Hölder inequality using associativity to get
$$\|f^2g\|_{L^1(\Omega)}=\|f\cdot (fg)\|_{L^1(\Omega)}\leq \|f\|_{L^\infty(\Omega)}\cdot \|fg\|_{L^1(\Omega)}\leq \|f\...
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Question about proof that normal operators have invariant subspace
Perhaps it is more transparent to apply the spectral theory of self adjoint operators. Assume $T$ is a normal operator and is is not a multiple of the identity operator. We have $T=A+iB$ where $A={1\...
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Variational formulation-exercise
The expression you want to prove relates to the inner product between functions in the Sobolev spaces H^(-1)(Ω) and H^1(Ω). To prove this, you'll need to use the properties of Sobolev spaces and ...
1
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Accepted
Eigenfunctions of 1D gaussian kernel
Taking the Fourier transform of both sides of
$$
\int_{-\infty}^{\infty}e^{-(x-y)^2/\sigma^2}f(y)\,dy = \lambda f(x), \tag{1}
$$
we obtain
$$
\sqrt{\pi}\sigma e^{-\frac{1}{4}\sigma^2k^2}F(k)=\lambda F(...
- 4,457
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Accepted
Continuous mean ergodic theorem
There are some assumptions missing in question (e.g. where f lives?, it is impossible to integrate a vector in an arbitrary Hilbert space), but this proof should work with these assumptions (if not ...
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1
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Accepted
Proving $\|fg\|_{L^1} \leq \|f\|_{L^p}^{\alpha} \|g\|_{L^q}^{1-\alpha}$
Let $f$ and $g$ be any positive functions for which all the integrals here are finite. Replacing $f$ by $nf$ we see that we must have $1 \leq \alpha$. Replacing $g$ by $ng$ we see that $1 \leq 1-\...
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Finding the conjugate of an operator between the Banach spaces $\ell_{p}$
The key to these kind of questions is how the duality is implemented. When we say that $\ell_q$ is the dual of $\ell_p$, what we mean is that any continuous linear functional $f:\ell_p\to\mathbb C$ is ...
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Why are coercive functions called coercive, and why is it useful?
I don't know if this can be helpful, but I'll provide an insight about the definition.
As already pointed out, a function $f:\mathbb{R^n} \rightarrow \mathbb{R}$ is usually said to be coercive "...
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Why are coercive functions called coercive, and why is it useful?
$\bf{Definition}$: A function $f : \mathbb{R}^n \to \mathbb{R^n}$ is coercive if and only if:
$$\lim_{\| x \| \to + \infty} \frac{f(x) \cdot x}{\| x \|} \to + \infty,$$
In terminology, this is named a ...
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Uniqueness of Sturm-Liouville like problem
I doubt that uniqueness holds under these assumptions. Consider any function $f\in C^1(\mathbb{R}, \mathbb{R})$ such that $f(\xi)=\pi^2\xi$ $(\xi \in [-1,1])$ and $f(\xi)=0$ $(|\xi| > 2)$. Clearly
$...
- 5,999
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Accepted
No nonzero continuous linear functional on $C([0,1])$ with this metric
As stated the assertion is trivially false because $V$ could be finite-dimensional, and the dual of a finite-dimensional TVS agrees with the algebraic dual and it has the same dimension as $V$.
The ...
- 199k
1
vote
Accepted
Functional Analysis trying to show that hilbert space equals $\sum \mathbb{F} e_j$
This statement is false. As has already been commented we can consider the sequence $x_n=\sum_{k=1}^n\frac{1}{k}e_k$. Since the sequence is cauchy it has a limit $x$ and it is easy to see that $<x,...
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Accepted
Doubt in a simple statement related to the maximal dyadic operator.
This is from the definition of supremum. The maximal function is the supremum of all averages $\frac1{|Q|}\int_Q |f(y)|dy$, so if the supremum is greater than $t$ there must exist a cube whose average ...
- 1,457
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On a locally-compact group $h(x^{-1})$ constant a.e. $\implies h(x)$ constant a.e. (proof of uniqueness of Haar measure)
Knapp's argument shows that there is a $\lambda$-null set $N\subset G$ such that if $x_0\in N^c=G\backslash N$ and we set $h_{x_0}(g) := |h(g^{-1}x_0)-h(x_0)|$ then $\text{supp}(h_{x_0}) = L_{x_0}$ ...
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Accepted
Are there any nonzero continuous multilinear functions $\ell^\infty(\mathbb{R})\to\mathbb{R}$?
No, even with just the second requirement. Assume $f(x) \neq 0$ for some $x$.
Let $y^n = \left((1 + \frac 11)\cdot x_1, (1 + \frac 12)\cdot x_2, (1 + \frac 13)\cdot x_3, \ldots, (1 + \frac 1n)\cdot ...
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Accepted
Trying to compute norm of an operator
You cannot expect to compute the norm of an operator of the form $T(h,k)=Th+Sk$ without consideration of the particular form of $T$ and $S$.
Consider $H_1=H_2=\mathbb C$, and $T=S=1$ and fix $h,k\in\...
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Convergence in weak topology implies pointwise convergence for a subsequence
This is not true. The sequence
$$
x_n := sign(\sin(n \pi t)
$$
converges weakly in $L^p(0,1)$ to $x=0$. But $x_n\ne0$ almost everywhere.
- 47.1k
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Does the "precise representative" $f^*$ uniquely represent classes in $L^1_{loc}$?
I tried to give myself an answer. However, the fastest way to see this is definitley, as it has been suggested in the comments, passing through Lebesgue-Besicovitch differentiation theorem, which ...
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Showing operator norm over $\mathbb{R}$ attains supremum
Consider the following set $S$={$v\in V;||v||=1$}$=B_V[0,1]$ by riesz compactness theorem $S$ is copmact since V is of finite dimension . And then define the mapping $N:S\rightarrow\mathbb R^+$ by ...
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Eigenvalues and eigenfunctions of the Fredholm integral equation of the second kind
Since
$$
Ku(x)=\sin(x)\int_{0}^{\pi}\sin(y)u(y)\,dy + \alpha \cos(x)\int_{0}^{\pi}\cos(y)u(y)\,dy, \tag{1}
$$
the eigenfunctions of $K$ associated with nonzero eigenvalues must be linear combinations ...
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Accepted
Extension of equivalent norm in subspace to the whole space
In my opinion the equality $B_{XX}\cap Y=B_{YY}$ suffices and there is no need to study the closure of $B_{XX}.$ Indeed,
for $y\in Y$ we have
$$\|y\|_{YY}=\inf\{r>0\,:\, y\in rB_{YY}\}=\inf\{r>0\...
- 25.1k
5
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Extension of equivalent norm in subspace to the whole space
[update: as I suspected this answer is more complicated than what you need (see Ryszard Szwarc‘s answer below for a simpler way to think about this), nonetheless here it is.]
We must show that $\text{...
- 4,002
2
votes
Accepted
Is the weak derivative a Radon-Nikodym derivative?
There is indeed a connection there, but your require a bit more regularity. You actually need $f\in W^{1,\infty}_{loc}(\mathbb{R})$, i.e. $f$ has to be locally Lipschitz. Then by Rademachers Theorem (...
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Fredholm Integral Equation of the Second Kind in $L_2[0,1]$
Let's first consider the simplest case: if $\lambda=0$, the solution to the integral equation
$$
x(t) + \lambda \int_{0}^{1}\left(\frac{1}{2} - |t-s|\right)x(s)\, ds = \cos(\pi t) \tag{1}
$$
is ...
- 4,457
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Gelfand - Mazur theorem
If $a \in \mathcal{A}$ and $a \neq 0$ then $\sigma(a) = \left\{\lambda \in \mathbb{C} : \lambda e - a \text{ is not invertible}\right\}$ is non empty (due to another theorem). Observe also that $\...
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Accepted
Is $a \mapsto(\pi(a) h \mid h)$ for $h \in \mathcal{H}$ with $\|h\|=1$ a state for non-unital C*algebra?
For the argument you are looking at, you don't really need to show that $$\tag1a\longmapsto\langle \pi(a)h,h\rangle$$ is a state; all that is needed is that it is a positive linear functional, because ...
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2
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Gelfand - Mazur theorem
An algebra $A$ in which the units (= invertible elements) are all non-zero elements of $A$ is called a division algebra. It is not true that any unital division algebra over $\mathbb C$ is isomorphic ...
- 2,501
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Gelfand - Mazur theorem
The proof seems fine to me. The assumption is that all non-zero elements of $\mathcal{A}$ are invertible. So,
$$a \neq 0 \implies \text{$a$ is invertible}. \tag{$\star$}$$
The contrapositive of this, ...
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