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$\sigma$-finite measure $L^p$ space is isometric to a finite measure $L^p$ space

This is a (very) late answer. The measure $λ$ given by $λ(Α) = \int_A h \, d\mu $ is finite and the operator $A \colon L^p(\mu) \to L^p(λ)$ given by $Af = h^{-1/p} f$ is an linear isometry since $$ \|...
Evangelopoulos Foivos's user avatar
3 votes
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Given $f \in L^p_{\text{loc}}$ and $\phi \in S$ (Schwartz class), do we have that $f \ast \phi \in C^\infty$? What if $f,\phi$ have compact support?

Claim 1 is false. If neither function has to have compact support, then the convolution may not be well-defined, e.g., take $\phi\in S$ nowhere zero, and $f=1/\phi$ which is continuous and therefore ...
Lieven's user avatar
  • 911
4 votes

Show $\|f\|_q\leq c\|f\|_r+\frac{1}{c^a}\|f\|_p$

I'm not sure about the exact value of $a$, but you can try this. Since $p < q < r$ there exists $0 < \lambda < 1$ satisfying $\dfrac 1 q = \dfrac \lambda p + \dfrac{1-\lambda}r$. Thus $$ 1 ...
Umberto P.'s user avatar
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1 vote
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Measurability of $\|f(\cdot, x_{2})\|_{L^\infty(X_{1})}$ (proof of Minkowski's inequality)

Thanks to @PhoemueX for the hint. We may use the $\sigma$-finiteness of $X_{1}$ to write $X_{1} = \bigcup_{n\geq 1}E_{n}$ with $\mu_{1}(E_{n}) < \infty$ and $E_{n}\subseteq E_{n+1}$ for all $n$. ...
Karthik Kannan's user avatar
2 votes
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Does convolution preserve local integrability in the sense that if $f \in L^p_{\text{loc}}$ and $g \in L^1$, then $f \ast g \in L^p_{\text{loc}}$?

As shown in the nice answer of daw, this is not true in this generality and I'd like to add another counterexample and a class of functions where it works. Another counterexample: For general $n\in \...
Severin Schraven's user avatar
1 vote

Solving integral: $\int_0^1\ln^2{\left(\frac{1+x}{1-x}\right)} dx$

In general, for any natural number $n$, we have $$ \boxed{\int_0^1 \ln ^n\left(\frac{1-x}{1+x}\right) d x=(-1)^n 2 n!\left(1-\frac{1}{2^{n-1}}\right) \zeta(n)} $$ Noticing the substitution $t=\frac{1-...
Lai's user avatar
  • 21.3k
1 vote
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Vector space generated by translates

Your deduction on $E$ is correct, so I will answer question 3. It seems it is talking about the vector space of $L^2$ functions defined on some interval $I\subset \mathbb R^n$ with values in the ...
Lorenzo Pompili's user avatar
3 votes

Does convolution preserve local integrability in the sense that if $f \in L^p_{\text{loc}}$ and $g \in L^1$, then $f \ast g \in L^p_{\text{loc}}$?

The convolution integral might not exist at all. Take $n=1$, $g(x) = min(1,|x|^{-3/2})$. Now let $f(x) = |x|^2$. Then $$ \int_{\mathbb R} f(x-y)g(y) = \int_{\mathbb R} |x-y|^2 g(y) dy\\ \ge\int_1^\...
daw's user avatar
  • 49.6k
3 votes
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Equivalent characterisation of the space $L^p_{\operatorname{loc}}(\Omega)$, where $\Omega$ is a non-empty open subset of $\mathbb R^n$?

Define $\Omega_\varepsilon:=\{x\in\Omega\,|\,\operatorname{dist}(x,\Omega^c)\geq\varepsilon,\text{ and }|x|\leq\varepsilon^{-1}\}$. You can verify that $\Omega_\varepsilon$ is compact, and it holds $$ ...
Lorenzo Pompili's user avatar
0 votes
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Confusion about Riesz representation theorem in $L^p$

For general $p$, this is far more difficult to construct than the representative in $L^2$ (which only needs the projection onto $\mathrm{ker}(T)$). In Alt - Linear Functional Analysis, this is done ...
Hyperbolic PDE friend's user avatar
2 votes
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Finding a function which is $L^1$ not $L^2$ and the integral is bounded by the square root.

We can simplify matters a bit. If the left side of the inequality is zero for some $A$, then the inequality holds trivially for such $A$. With this in mind, we can assume $f \equiv 0$ on $(-\infty, 0]$...
stoic-santiago's user avatar
5 votes
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Notation $D^{m}$ in certain inequalities in $L^{p}$

If you look into the original paper of Nirenberg (see here; p. 125), it is stated that $\Vert D^{l}u\Vert_{L^{p}}$ for $l\in\mathbb{N}$ is defined by $$\Vert D^{l}u\Vert_{L^{p}}:=\max_{\vert\alpha\...
G. Blaickner's user avatar
4 votes
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Prove that Fourier transform is $L^p$

If you can show $f(x)$ is in $L^1({\mathbb R}^n)$, then $\hat{f}(\xi)$ is in $L^{\infty}({\mathbb R}^n)$, and if you can show $f(x)$ is in $L^2({\mathbb R}^n)$, then $\hat{f}(\xi)$ is in $L^{2}({\...
Zarrax's user avatar
  • 45.3k
0 votes

Set of sequences with zero sum density in $l_2$

We denote by $c_{00}$ the subspace of sequences that are eventually zero, and $c_0$ the space of sequences that converge to zero, which is a closed proper subspace of $\ell^\infty$. Then $c_{00}$ is ...
hbghlyj's user avatar
  • 2,812
1 vote
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How pathological are the $L^{p}$ functions?

There exists an $F \in L^1([0,1])$ unbounded in every subinterval. Proof: Let $\varphi^{a}_n: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by: \begin{align*} \varphi^a_n(x) = \...
Rohan Didmishe's user avatar
2 votes
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A sequence $(x^k) \subset l^p$ with $|x^k_n| \leq a_n$ for $(a_n)\in l^p$ has a convergent subsequence

The question has some answers. I propose a different approach based on compact operators. Consider the operator $T:\ell^\infty \to \ell^p$ defined by $(Tx)_k=a_kx_k.$ Let $T_n:\ell^\infty \to \ell^p$ ...
Ryszard Szwarc's user avatar
2 votes
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$X \approx \ell^p(I,X)$ $\Rightarrow$ $X \approx X \oplus X$ (used in The Pełczynski decomposition technique)

If $X$ is a vector space and if both $X_{1}$ and $X_{2}$ are vector subspaces over $X$ such that $X_{1} \cap X_{2} = \{0\}$, then it is possible to form another vector subspace which consists of all ...
Dean Miller's user avatar
  • 1,593
0 votes

Real Analysis, Folland Problem 6.1.2 $L^p$ spaces

Part (a) second claim's statement from @NapD.Lover is not exactly correct, since $|g| = \|g\|_\infty$ need not hold for all $x \in \{x \in X : f(x) \neq 0 \}$; it only need to hold for a.e. $x$ such ...
Squirrel-Power's user avatar
0 votes
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Are Sobolev–Hölder functions continuous up to the boundary?

Yes, that is implied. If you look into proofs of these embedding theorems, you can see this: First, the claim is proven for smooth functions. Then using density arguments the claim is proven for ...
daw's user avatar
  • 49.6k
5 votes
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Sandwiching the Lp norm sequence of random variable

The claim is false. Here is an explicit counterexample. Let $X$ be a non-negative random variable with density given by $$p(x) = \sum_{k=1}^\infty \frac{1}{2^k}C_{3/2+1/2k} \exp\left ( -x^{3/2+1/2k} \...
Matt Werenski's user avatar
0 votes

Rademacher functions form an orthonormal system but not an orthonormal basis

The system is complete but it's not a basis because it misses constant functions, i.e. $<1,r_n>=0$ for all n. In fact, the $L^2$ closure of the span of the $r_n$'s is also closed in the weaker ...
user610395's user avatar
3 votes
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A property of solution operator of a elliptic PDE involving positive part of a function

I think this inequality cannot be true. Here is a counterexample for $N=1$: Define $u_n(x) := \sin(2n\pi x) \chi_{[0,1]}$. Then $u_n \rightharpoonup 0$ in $L^2(0,1)$ and in $L^2(\mathbb R)$. The ...
daw's user avatar
  • 49.6k
1 vote
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Showing $L^2$ - convergence of $\phi_n(X_n)$ to $\phi(X)$ when $\phi_n \to \phi$ and $X_n\to X$

This is not necessarily the case. Assume your probability space is the standard $((0,1), \mathcal L)$ ($\mathcal L$ is the lebesgue measure) and define the random variables $X_n$ as follows: $$X_n(x):=...
Pelota's user avatar
  • 843

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