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0

The $1$ in parentheses is useless; it doesn't scale if you multiply $u$ by a constant. So your inequality can hold for all $u\in H^1$ if and only if $$\|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{\frac{p}{2}})+C_\epsilon \|u\|_2$$ holds. And here the scaling of the variable $u_t = u^{n/2^*}u(tx)$ becomes an issue: namely, $\|u_t\|_{2^*}$ and $\|\nabla ...


0

You have that if $a \in l_q\big(\Bbb N\big)$ then $a \in l_q\big(\Bbb Z\big)$ since a sequence of natural numbers is a sequence of integers. You also have if $a$ and $b$ are two sequences of natural numbers then so is $ a*b$, since the convolution only involves multiplications and additions. So the equality you wrote holds for sequences of natural numbers ...


2

I can't bring myself to write $fg$ for the convolution of $f$ and $g$. So I'm going to write $f\mapsto f'$ for the involution, so I can write $f*g$ for the convolution. Are you certain you got the definition of $f'$ straight? What would make much more sense to me would be $$f'(t)=\overline{f(-t).}$$ That seems to me is the "standard" involution on $L^1$. ...


2

For any sequence $f_n \to f''$ in $L^p$, there exists a subsequence such that $f_{n(k)} \to f''$ almost everywhere as $k \to \infty$. There are (at least) two possibilites to prove this statement: In the proof of the Riesz-Fischer theorem (which states that $L^p$ is a complete space), one usually constructs such a sequence (see e.g. René Schilling: ...


0

Note that if $\mu(X)=\infty$, then your series condition doesn't necessarily imply $f\in L_{\mu}^{p}(X)$. For example, take $X=\mathbb{R}$ with Borel $\sigma$-algebra and Lebesgue measure. If $f=1/2\chi_{[0,\infty)}$, then clearly $f\notin L^{p}$; however, $\lambda_{f}(n)=0$ for $n\geq 1$. Assume that $\lambda_{f}(0)<\infty$. Observe that for $p>0$, ...


1

It holds true that $q>p\implies \ell^p\subseteq \ell^q$ Indeed, let $(a_n)_{n\in\mathbb N}$ such that $\sum_n |a_n|^p<+\infty$ Since $q>p$, as soon as $|a_n|<1$, which holds from a certain $N$ onwards, $|a_n|^q<|a_n|^p$. So $\sum_{n\in\mathbb ...


1

Fix $y \in [0,\infty)$. Let $f_N(x,y) := \sum_{k=1}^N e^{-y\sqrt{\lambda_k}} (u,\varphi_k) \varphi_k(x)$ be the partial sum, and let $g_N := f-f_N$ be the tail of the series. Consider the integral$\int_{\Omega } g_N^2$: writing $g_N^2$ as a product of two sums and multiplying out the terms, then only the diagonal terms survive after integrating because the ...


4

No, this is not correct. Just consider $((0,1],\mathcal{B}((0,1]),\text{Leb})$ and $$X_n(\omega) := n^2 \cdot 1_{(0,1/n)}(\omega), \qquad \omega \in (0,1].$$ Then $X_n \to 0$ almost surely (hence, in particular, $X_n \to 0$ in probability), but $\mathbb{E}X_n = n$ is not bounded. Another counterexample: $$X_n(\omega) := \frac{1}{\omega} ...


0

Split into small $x$ and large $x$: $|x|\le 2$. Nothing to discuss here: the function is continuous. $|x|>2$. Then $u(x)\le \dfrac{1}{|x| \ln^p |x|}$. By symmetry, it's enough to consider positive $x$. Integrate using substitution $y=1/x$. This is one of standard improper integrals that just barely converge (thanks to $p>1$).


1

Hint: Write $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$. By Young's inequality, $$|u_i \cdot v_i| \leq \frac{|u_i|^p}{p} + \frac{|v_i|^q}{q}. \tag{1}$$ Rewrite this inequality using the definition of $u$ and $v$. Sum $(1)$ over $i=1,\ldots,n$. Deduce that $$\sum_{i=1}^n |u_i v_i| \leq 1.$$ Conclude.


1

This is more of a problem relating to weak convergence in $L^{2}$. Since $f_{n}\to f$ weakly in $H^{1}\left(\Omega\right)$, we know $g_{n}:=\nabla f_{n}\to\nabla f=:g$ weakly in $L^{2}\left(\Omega\right)$. Now, we require the following fact: If $x_{n}\rightharpoonup x$ weakly and $y_{n}\to y$ in norm, then $\left(x_{n}\right)_{n}$ is bounded (lets say by ...


1

I think this does it: if a sequence converges in $L^1$, then a subsequence converges a.s. to the same limit. That subsequence converges in probability by Egorov's theorem. So the $L^1$ limit of an $L^1$-convergent sequence must be in an $L^0$-closed set. The only catch here is that this assumes that you are dealing with a subset of $L^1$ which is closed in ...


0

HINT: Once you have a continuous $g$ that is a good approximation to the generic $f\in L^2(-1, 1)$, take a $\delta>0$ and do a linear fit between the points $(-1, \alpha)$ and $(-1+\delta, g(-1+\delta))$. That is, consider the function $$ g_\delta(x)=\begin{cases} \text{linear}, & x\in[-1, -1+\delta) \\ g(x), & x\in [-1+\delta, 1] \end{cases} $$ ...


1

Generally, if $T:X\to Y$ is a surjective linear map, then $X/\ker T$ is isomorphic to $Y$. Applied to your case, $X=L^p$, $Y=\mathbb{R}$, $T=\int f$, this yields $L^p/L^p_0 $ being isomorphic to $\mathbb{R}$. (Or $\mathbb{C}$ if you use complex scalars) Bonus content: If $M$ is a closed subspace of Banach space $X$, the inclusion $i:M\to X$ has adjoint ...


0

Writing $l(x)$ for $|\log|x||$. Hints: First, all that matters is what happens for, say, $|x| > 10$ and what happens for $|x|<1/10$. For $|x|>10$, $1+|x|^\alpha \sim |x|^\alpha$ and $1+l(x)^\beta\sim l(x)^\beta$. After that "substitution", write the integral of the $p$-th power in polar corrdinates; you get $\int_{10}^\infty t^\gamma ...


2

Let me give you a hint: First, let us define $$v_\epsilon:= \epsilon^{-\frac{n}{p^*}}u\left(\frac{x}{\epsilon}\right) $$ Can you compute $\|v_\epsilon\|_{L^p}$ and $\|\nabla v_\epsilon \|_{L^p}$ in term of $u$? Try to write it down explicitly. Then you will know why it is bounded in $W^{1,p}$ Secondly, the fact that $v_\epsilon$ has no convergent ...


1

Well first there's a typo in the question; you mean to ask how to prove $||f*g||_p \le||g||_p$, right? And there's some funny business in the question itself; smoothness and compact support are irrelevant. (And they don't make it any easier as far as I can see.) All that matters is $f\ge 0$ and $\int f=1$. That means if you define a measure $\mu$ by ...


1

The first question is, in what sense do you want to understand convergence of the sum? The correct notion here is unconditional convergence, i.e. we show that there is some $h \in H$ such that for every $\epsilon > 0$, there is some finite subset $J_\epsilon \subset I$ with $\Vert h - \sum_{j\in J} a_j x_j \Vert < \epsilon$ for all finite sets ...


2

I am not sure about how $\tau$ is a translation. But the existence of the required functional $\Psi$ can be proved in a more general setting. It will cover your case if we assume that $\{nq-1\}_{n=1}^\infty$ is a strictly increasing sequence of positive integers. This is my proof: Consider any strictly increasing function ...


-1

Looks like a job for the Schauder fixed-point theorem. EDIT: I'm assuming $q$ is an integer > 1, so $n \in \mathbb N \implies qn - 1 \in \mathbb N$. The unit ball $B$ of $(\ell^\infty)^*$ with the weak-* topology is compact. The intersection $K$ of the subsets $\{\psi \in B: \liminf x \le \psi(x) \le \limsup x \}$ is again compact and convex (and ...


0

It seems the following. I will consider that $s\in\Bbb R$, not in $\Bbb R^n$, because I don’t understand what is $sx\in\Bbb R^n$ The correctness of the definition of the operator imposes restrictions on the set $U$: $U+a\subset U$ or $sU\subset U$. Or we may for each function $f\in L^p(U)$ consider its extension in $L^p(\Bbb R^n)$ such that $f|(\Bbb ...


2

Use that $\chi_{(a,b]} = \chi_{(0,b]} - \chi_{(0,a]}$ for $a<b$. This will show that your first assumption yields $$ \int f_n \cdot \chi_{(a,b]} \to \int f \cdot \chi_{(a,b]} $$ for all $a<b$. By linearity, we get the same convergence for arbitrary Riemann step functions, i.e. for step functions with respect to intervals. Now, one possibility is to ...


2

Let $S\equiv\sup_{x\in\mathbb R^n}|f(x)|$. By definition, the essential supremum norm is defined as follows: $$\|f\|_{\infty}=\inf_{c\geq 0}\big\{\lambda(\{x\in\mathbb R^n\,|\,|f(x)|>c\})=0\big\}.$$ In words, $\|f\|_{\infty}$ is the infimum of such non-negative numbers above which the function $|f|$ takes values only on a measure-zero set. Intuitively, ...


1

Since the Lebesgue measure is $\sigma$-finite we have that $$\| {f}\|_{\infty}=\inf\left\{M\geq 0:\,\lambda(\left\{x: |f(x)| > M\right\})=0 \right\}$$ Clearly $\lambda(\left\{x: |f(x)| > \sup_{x \in \mathbb{R}^{d}}|f(x)|\right\})=0$ Now suppose there is a constant $M < \sup_{x \in \mathbb{R}^{d}}|f(x)|$, that satisfies the above critera. It should ...


0

For c), apply Fatou's lemma to $2^{p-1}(|f|^p + |f_k|^p) - |f - f_k|^p$, giving $$\liminf_{k \to \infty} \int_X 2^{p-1}(|f|^p + |f_k|^p - |f - f_k|^p)\, d\mu\ge 2^p \int_X |f|^p\, d\mu.$$ Thus $$2^{p-1}(\|f\|_p^p + \liminf_{k\to \infty} \|f_k\|_p^p) - \limsup_{k\to \infty} \|f - f_k\|_p^p \ge 2^p\|f\|_p^p.$$ Since $\|f_k\|_p^p \to \|f\|_p^p$, we have ...


2

You haven't got the correct negation of the limit hypothesis. Remember that the opposite of $f(x) \to b$ as $x \to \infty$ if for every $\varepsilon>0$ there is an $n$ such that $\lvert f(x)-b \rvert < \varepsilon$ for all $x>n$ is $f(x) \not\to b$ as $x \to a$ if there is an $\varepsilon>0$ such that for every $n$ there is an $x>n$ ...


3

Use Holder to see $$(1)\,\,\,\,|\int_x^{x+1}f\,| \le \int_x^{x+1}|f| = \int_x^{x+1}|f|\cdot 1 \le (\int_x^{x+1}|f|^p)^{1/p}\cdot (\int_x^{x+1}1)^{1/q}=(\int_x^{x+1}|f|^p)^{1/p}.$$ Because $f\in L^p, \int_x^{x+1}|f|^p \to 0$ by the dominated convergence theorem. That shows $(1) \to 0$ as desired.


4

Thanks to zhw for catching an oversight. Write $\mathbb{R} = \bigcup_n [n,n+1)$ and then $$\int_\mathbb{R} |f|^p = \sum_n \int_{[n,n+1)} |f|^p,$$ and hence $$\int_{[n,n+1)} |f|^p \stackrel{n\to\infty}{\longrightarrow} 0.$$ If $p=1$, we are finished, otherwise suppose $p>1$. Since $$\left(\int_{[a,b)} |f|\right)^p = \left(\int_{[a,b)} |f|1\right)^p \le ...


1

The idea to use $f_k=k^\alpha \chi_{(0,k^{-1})}$ is good. The sequence converges to $0$ a.e., so if there is any $L^p$ limit at all, it has to be zero (recall that $L^p$ convergence implies a.e. convergence for a subsequence). Therefore, if the sequence of norms stays constant, the sequence does not have a limit in $L^p$. Choose $\alpha$ so that these ...


1

This was getting too long for a comment. Analysis is not my thing, but I think you've reduced the problem to showing that for a continuous function $f\in \mathcal{C}^0(\left[0,1\right])$ and $\epsilon > 0$, you can find $g\in E_\alpha$ such that $\left|f-g\right|_2<\epsilon$. So, given a continuous $f$, let $g$ be equal to $f$ except on a super tiny ...


2

This follows from the triangle inequality of the$\ell^2$ norm, since $x,y \in \ell^2$ then $\|x\|_2 <\infty $ and $\|y\|_2 <\infty$ thus $$ \sum_{n=0}^\infty |x_n + y_n|^2 =\| x+ y\|_2^2 \leq (\|x\|_2+\|y\|_2)^2 < \infty $$ Hence, indeed $x+y \in \ell^2$


3

Absolutely cloddish inequality: If $a,b \ge 0,$ then $(a+b)^2 \le 4a^2 + 4 b^2.$ Proof: If $a\le b,$ then the left side is $\le (2b)^2 = 4b^2,$ same idea of course if $a\ge b.$ So $$\sum (x_n+y_n)^2 \le \sum (|x_n|+|y_n|)^2 \le \sum (4|x_n|^2 + 4|y_n|^2)$$ and that does it.


1

Hint: multiply out the terms in the sum for $||x + y||$, and use Cauchy-Schwarz to find a bound for $\sum x_iy_i$ in terms of $||x||$ and $||y||$.


2

$(\Rightarrow)$ By the reverse triangle inequality $$ | \|g_n\| -\|g\| | \leq \| g_n -g \| $$ Since $\| g_n -g \| \to 0$ then clearly $\|g_n\| \to \|g\|$ The converse is also true as long $g, g_n \in L ^1$ and $g_n \to g$ a.e. In that case you can use the next argument: $(\Leftarrow)$ First note that since $|g_n -g| \leq |g_n| + |g|$ then $|g_n -g| ...


0

It looks like your reasoning is fine, except that you may want to add some more detail as to how $$\left|\, \int_{X} g_n \, dm - \int_{X}g\, dm \,\right|$$ is related to the (absolute value of) the difference of the norms. Specifically, I’d write something like $$\left|\, ||g_n||_{1} - ||g||_1 \,\right| = \left|\, \int_{X} |g_n| \, dm - \int_{X} |g| \, dm ...


0

In the case $p = 2$, we first compute $T^*$. Noting that $$ \langle Tx, y\rangle = \sum_{k} \sum_{j = 1}^k x_j \bar{y}_k / k = \sum_{j} \sum_{k = j}^\infty x_j \bar{y}_k / k $$ we have that $$ (T^*y)_j = \sum_{k = j}^\infty \frac{y_k}{k} $$ So we have that $$ (T T^* y)_\ell = \sum_{j = 1}^\ell \sum_{k = j}^\infty \frac{y_k}{k \ell} $$ and $$ (T^* T ...


5

The dual space of $L^\infty$ is the space of "finitely addditive" signed measures and not the space of finite signed measures. The usual Radon-Nikodym theorem does not apply.


0

The $\xi$-integral can be computed as $$ \int e^{i(x-y)\xi}e^{-|\xi|^2/n}\mathrm{d}\xi = e^{-n|x-y|^2}\int e^{-(\xi/\sqrt{n}-i\sqrt{n}(x-y))^2}\mathrm{d}\xi = (\pi n)^{3/2} e^{-n|x-y|^2} , $$ and hence $$ I_n(x)=\int\chi_{B(n)}(y)f(y)(\pi n)^{3/2} e^{-n|x-y|^2}\mathrm{d}y = \int f_n(y)g_n(x-y)\mathrm{d}y , $$ where $$ f_n(y) = \chi_{B(n)}(y)f(y) , $$ and ...


-1

Let $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \sin (2\pi i \,n x) + b_n \cos (2\pi i \,n x) \in L^2[0,1]$ then we have Bessel inequality: $$ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) < \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 < \infty$$ Then the Fourier coefficients tend to zero $|a_n| =|\langle f ,\sin (2\pi i \,n x)\rangle | \to ...


0

Suppose that $\mu(B_{n})=0$. That would imply that the $a$ in the definition of $\Vert f\Vert_{\infty}$ satisfies $a\leq \Vert f\Vert_{\infty}-\frac{1}{n}$. Since we assumed that $\mu(\{|f(x)|>\Vert f\Vert_{\infty}-1/n\})=0$. But this is non-sense since it implies \begin{align*} \Vert f \Vert_{\infty} &\leq \Vert f\Vert_{\infty}-\frac{1}{n} \\ ...


1

If both $f$ and $g$ are continuous with compact support, then so is their convolution. And continuous functions with compact support are dense in $L^p$ for $1<p<\infty$. So, you need to do the following: Given $f\in L^p$ and $g\in L^q$, approximate them with $C_c$ functions $f_n$ and $g_n$ in the respective norm. Show that $f_n*g_n$ is a Cauchy ...


0

Indeed this is just Holder's inequality: Pick $V \subset U$ with $|V|<\infty$, then $$||u||_{L^1(V)}=\int_V |u| dx\le \sqrt{\int_V 1^2 dx}\sqrt{\int_V |u|^2 dx} = \sqrt{|V|}\ ||u||_{L^2(U)}$$ then $u \in L^1_{loc}(U)$.


1

Try $X = \mathbb R$ with the usual topology, and counting measure on the rationals. The only continuous function in $L^1$ is $0$.


2

For the case $1\leq p_1 < p_2 < \infty$, let $p=p_2/p_1 >1$. Then take any $f \in L^{p_2}(E)$, this means $$ \int_{E} |f|^{p_2} dm < \infty $$ Now, using that $p=p_2/p_1$ we observe that $$ \int_{E} |f^{p_1}|^{p} dm = \int_{E} |f|^{ p_1 \cdot p} dm =\int_{E} |f|^{p_2} dm < \infty $$ Thus indeed $f^{p_1} \in L^p(E)$. Now for the rest, note ...


1

Given $f \in L^{p_2}$, $f^{p_2} \in L^1$. So since $p_2 = p_1p$, $(f^{p_1})^p\in L^1$, i.e., $f^{p_1}\in L^p$.


2

Let $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. If $X$ is a set, $\mathcal{S}$ a sigma algebra on $X$ and $\mu : \mathcal{S} \to \mathbb{R}^+$ a measure, then, by definition $$ L^p(X, \mathcal{S}, \mu) = \left\{ f:X\to \mathbb{F} : f \text{ is $\mathcal{S}$ measurable and } \int_X |f|^p d\mu < \infty \right\} $$ In that case, if $f \in L^p(X, \mathcal{S}, ...


0

Actually, $f$ is not necessarily bounded. $f(x)=\frac{\chi_{(0,1]}(x)}{\sqrt{x}}$ is an unbounded (Lebesgue) integrable function (its integral is $2$). Hint: try to argue that $\int_{-\infty}^n f(x) dx$ converges to $\int_{-\infty}^\infty f(x) dx$ by using the dominated convergence theorem. Then notice that $\int_{-\infty}^\infty f(x) dx - \int_{-\infty}^n ...


0

Well the part $\int_{-\infty}^nf(x)dx$ must converge to $\int_{-\infty}^{\infty}f(x)dx$ so the part $\int_n^{-\infty}f(x)dx$ must go to zero. Just write the whole integral as a sum of those two parts and take the limit.


1

This is true provided that $C_b \cap L^1$ is dense in $L^1$, which holds if $m$ is a finite measure, or if $X$ is locally compact and $m$ is Radon (I imagine many other conditions are possible). In this case, we can use a standard triangle inequality trick. Given any $h \in L^1$ and any $\epsilon > 0$ we can find $g \in C_b \cap L_1$ such that ...


0

Even in the finite measure case this seems false. Take $f_n(x)=\mathrm{sign}(2^n x)$ in $L^\infty(0,2\pi)$. Then for all $n \neq m$, $\mathrm{Leb}\{ x , |f_n(x)-f_m(x)|>1\}=\pi$.



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