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5 votes
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Limit of integral of sum of cosine functions by CLT?

By symmetry, rewrite the integral as $$\lim_{n\to\infty}\frac{(\sqrt{n})^d}{\pi^dd^{2n}}\int_{[0,\pi]^d}\mathrm{d}^d\mathbf{x}\left(\cos(x_1)+\cdots+\cos(x_d)\right)^{2n}$$ $$ = \lim_{n\to\infty}\frac{...
Ninad Munshi's user avatar
  • 35.2k
4 votes
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Analysing a Lebesgue integral inequality for $|t^{-n} \phi(x/t)|$, where $\phi \in C_c^\infty \cap L^1$ with $\| \phi \|_1 = 1$.

The essential ingredients are that $\phi\in L^1(\mathbb{R}^n)$ and that $\phi$ has compact support (no need to assume any regularity or boundedness). The compact support condition is telling us that ...
Severin Schraven's user avatar
4 votes

Analysing a Lebesgue integral inequality for $|t^{-n} \phi(x/t)|$, where $\phi \in C_c^\infty \cap L^1$ with $\| \phi \|_1 = 1$.

Since $\phi \in C_{c}(\mathbb{R}^{n})$, there is $R > 0$ such that $\phi(x) = 0$ if $|x| \geq R$, so $\phi_t(z) = 0$ if $|z| \geq tR$, now, using the change of coordinates $z = ty$ you have $$ \...
Raul Fernandes Horta's user avatar
4 votes

$\left |\int _{X} log(|f|)d\mu \right | < \infty$ if $\int_{X} |f| d\mu <\infty $?

The integral may diverge for two reasons. The first one concerns the measure by which we integrate. If the measure is large, the integral may also be large (like $\int_{\mathbb{R}}1 \,\mathrm{d}\...
Marek Kryspin's user avatar
4 votes
Accepted

$e^{-\lVert x \rVert ^2}$ is integrable over $\mathbb{R}^n$ and $\int_{\mathbb{R}^n} e^{-\lVert x \rVert ^2} = \pi^{n/2}$

$$\int\dots \int e^{-\sum_i x_i^2} \prod_k \ dx_k =\int \prod _i \ e^{-x_i^2} dx_i = \prod _i\int e^{-x_i^2} dx_i.$$ The integral may be splitted into a power of the square $$\left(\int e^{-x^2-y^2} ...
Roland F's user avatar
  • 2,683
3 votes

Confusing different definitions of Lebesgue Integral of Simple Functions

A given simple function can have many representations, as you have noted. But, there is at least one representation using disjoint sets, sometimes called a canonical form of the simple function. It ...
Jürgen Sukumaran's user avatar
3 votes
Accepted

Lebesgue-Vitali theorem and the characteristic function of $\mathbb{Q}$

Almost everywhere continuity means that the set of points where the function is not continuous has measure zero. The indicator of the rationals is continuous nowhere, because the rationals are a dense ...
kieransquared'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
2 votes
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Is this a typo in Gilbarg Trudinger?

In G&T, $\omega_n= \vert B_1\vert$. Hence, $\vert \partial B_1\vert =n \omega_n $, so the extra $n$ cancels with the $n$ on the denominator.
JackT's user avatar
  • 7,159
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
2 votes

$f:[0,1] \to R$ is measurable but not Lebesgue integrable. Is $g(x,y) := f(x) - f(y)$ is also not Lebesgue integrable on $[0, 1] × [0, 1]$?

If $f$ is not integrable, then neither is $f-\alpha$ for any constant $\alpha$ since the measure of the whole space is finite. In particular, for any $y$, $\int |f(x)-f(y)| dm(x) = \infty$ and hence $...
copper.hat's user avatar
  • 173k
2 votes
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If $\phi \in C_c^\infty(\mathbb R^n) \cap L^1(\mathbb R^n)$ such that $\| \phi \|_1 = 1$, then the dillations $\phi_t$ preserve these properties?

Note that $$t\{u\in\mathbb{R}^n:\phi(u)\neq0\}=\{tu\in\mathbb{R}^n:\phi(u)\neq0\}=\left\{x\in\mathbb{R}^n:\phi\left(\frac{x}{t}\right)\neq0\right\},$$ from which it easily follows that $$\operatorname{...
Lorago's user avatar
  • 10.1k
1 vote
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On surface of $C = \{ x^2+y^2=1,0<z<1 \}$ there is subset A, $A_t=A \cap \{z <t \}$. Prove $\int_0^1 \lambda_2(A_t) dt= \iint_A (1-z)d \lambda_2$

Let $a_{t}={d\lambda_2(A_t)}/{dt}$ represent the 1-dimensional volume of the part of $A$ at height $t$ (defined for almost every $t\in[0,1]$). Then the left hand side is equal to $$\int_{t=0}^1\int_{z=...
Lieven's user avatar
  • 921
1 vote

"Leibniz's rule" for $t\in\mathbb{R}^n$

Partial derivatives are derivatives of single-variable functions obtained by fixing all other points, so the proof is easily reduced to the case $n=1$. If you want a reference, look at Amann Escher, ...
peek-a-boo's user avatar
  • 57.3k
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
1 vote
Accepted

Finiteness of the integral $\int_{\mathbb{R}} |f| \ln(|f|) dx $

Usually, when we bring up an integral, like $\int_\mathbb{R} |f| \ln(|f|)\,dx$ and put it in an inequality as an assumption, that means that we're assuming the integral exists. In the context of the ...
Brian Moehring's user avatar
1 vote
Accepted

How to bound $\mu (\{ x \le X : |ψ(x) − x| \ge εx^{1/2} (\log x)^2 \})$ from above?

Here's a different estimate that you can prove. Via a dyadic subdivision, $$\mu\left(\left\{2 \leq x \leq X : |\psi(x) - x| \geq \varepsilon \sqrt{x}(\log x)^2\right\}\right) \leq \sum_{\ell = 2}^{\...
Peter Humphries's user avatar
1 vote
Accepted

Show that $\int_a^bF(x)dx+\int_{F(a)}^{F(b)}F^{-1}dy=bF(b)-aF(a)$

Your initial idea is good, we can use a substitution with Lebesgue integrals in the following, if $g \colon X \to Y$ is a measurable bijection, then (you can prove this for characteristic functions ...
Raul Fernandes Horta's user avatar
1 vote
Accepted

General version of the fundamental lemma of calculus of variations

Remark: Here, we are assuming that $\mu$ is locally finite, such that mollified functions convergence pointwise a.e. to the original function. This is a standard argument, that can probably be found ...
Hyperbolic PDE friend's user avatar
1 vote
Accepted

Convergence of the difference of the convolution of sequence of functions and a function

Let $\epsilon > 0$ then there exists $\delta>0$ s.t. $\Vert \tau_af-f\Vert_1<\epsilon$ for all $\Vert a \Vert < \delta$. Furthermore, there exists $N\in\mathbb{N}$ such that if $j\geq N$ ...
Mathematician....'s user avatar

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