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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-\...
geetha290krm's user avatar
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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.
daw's user avatar
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2 votes
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An inequlaity involving an $L^p$ term

Without losing the generality, assume $A \geq B > 0$ and set $r = \frac{B}{A} \in (0, 1]$. Then consider the function $$ f(p) = \int_{0}^{2\pi} |1 + re^{i\theta}|^p \, \mathrm{d}\theta. $$ Since $p ...
Sangchul Lee's user avatar
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Bounding an integral by a Besov norm

The ingredient you're missing is Minkowski's integral inequality, which lets you move the $\frac32$ inside the integral and interchanges the order of integration. This gives \begin{align*} \lVert I_{\...
ktoi's user avatar
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1 vote

How to prove that $\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}\in L^1(\mathbb R^{2n})$?

Ad 1: As $u_n\rightarrow u$ weakly in $X^{s,p}(\Omega)$, we have $u\in X^{s,p}(\Omega)$. Ad 2: Considering the function $f:\mathbb{R}^{2n}\rightarrow\mathbb{R}$, $(x,y)\mapsto \frac{\vert u(x)-u(y) \...
stange's user avatar
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$\int \lvert f_n-f\rvert^p g\rightarrow 0$ implies convergence in measure?

Counter-example: $\Omega =(-1,1),f(x)=0, f_n(x)=x^{2} $ for all $n$ and let $g(x)$ be any odd function with compact support, not identically $0$.
geetha290krm's user avatar
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2 votes
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Is an absolutely bounded function in L1 also in L2?

This follows from Holder's inequality: $$ \| f \|_2^2 = \| f^2 \|_1 \leq \| f \|_1 \| f \|_\infty < \infty $$
Jose Avilez's user avatar
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1 vote

Help with the last step in proving Hardy–Littlewood maximal inequality

This is a VERY back-of-the-envelope calculation. But I think the general idea is to use Fubini theorem to flip the order of integration:
Scott Hahn's user avatar
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1 vote

"Scaled $L^p$ norm" and geometric mean

Just as a remark, these are the unit-circles with respect to the $L^p$ norm (not a real norm for $p<1$): These can be obtained by using polar coordinates $x=r⋅\cosφ$ and $y=r⋅\sinφ$, since then $$‖...
Hyperplane's user avatar
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Proof by contradiction that $(\ell^p)^* \subseteq \ell^q$

Nick's idea works, for the first question at least. We let: $$\mathbf{x}^n = (y_1^{q-1}, y_2^{q-1}, \ldots, y_n^{q-1}, 0, 0, \ldots),$$ and define $$\mathbf{u}^n = \frac{\mathbf{x}^n}{\|\mathbf{x}^n\|...
Theo Bendit's user avatar
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2 votes
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Brezis' Lemma 8.1: where did I get wrong?

$\varphi$ is not necessarily compactly supported, as to have that condition on $\phi$ you need to have the following: $$0 = \varphi(b) - \varphi(a) = \int_I \varphi' = \int_I \psi$$ which you did not ...
Bruno B's user avatar
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4 votes
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Can $\mathbb{R}$ be expressed as a disjoint union of two dense locally non-zero measure measurable subsets?

Yes, such a decomposition exists. (*update: As Nate Eldredge points out in the comments, your motivating question on $L^\infty$ denseness of piecewise constant functions is already contradicted by a ...
M W's user avatar
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1 vote

Must the domain of a Hilbert space operator be limited to square-summable elements? If so, what type of space does not restrict, thus?

The so-called delta function, formally, is a tempered distribution where the tempered distributions (denoted $\mathcal S'$) are defined to be the dual of Schwartz space (denoted $\mathcal S$), i.e. $\...
Frederik vom Ende's user avatar
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Relation of $L^1$ and "$L^{1/2}$"

The question, pre-edit, was also true, though it took a bit more work. The current problem may be answered simply: Note that when $0 < \delta < \pi/4$, your $h$ satisfies $$0 < A := \left(\...
Brian Moehring's user avatar

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