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2

If you know that $C_0^\infty((0,1))$ is dense in $L^2([0,1])$, a natural approach is to do what you did: take a sequence $\phi_n$ in $C_0^\infty((0,1))$ converging to $f$ in $L^2$ and obtain, $$ \int_0^1 |f|^2 = \int_0^1 (f-\phi_n)\overline{f} \le \|f-\phi_n\|_{L^2}\|f\|_{L^2}\to 0 $$ hence $\int_0^1 |f|^2=0$. Without the density result, I would use a ...


1

As suggested in the comments, the sequence of partial sums $g_k = \sum_{n=1}^k |f_{n+1} - f_n|$ is monotonically increasing and converges to $g$. Then it follows from Monotone Convergence Theorem that $\lim \int g_k = \int g$ and similarly $\lim ||g_k||_p = \int ||g||_p$. Applying triangle inequality and $||f_{n+1} - f_n||_p < 2^{-n}$, you can show each ...


0

The inequality follows from the fact that if $a$ and $b$ are two real numbers, then (by convexity of $x\mapsto |x|^p$, $$|a-b|^p\leqslant (|a|+|b|)^p\leqslant 2^{p-1}(|a|^p+|b|^p),$$ hence by conditional Jensen's inequality, $$|\mathbb{E}[X_\infty1_{|X_\infty|> K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|> K}|^p \leqslant 2^{p-1}\left(\mathbb ...


1

$p=2$ does the job. To see this, represent the bounded linear form $\Phi$ as $\Phi(x)=\langle x_0,x \rangle$. Bessel's equality gives the wanted convergence. Indeed, we have $$\sum_{n=1} ^\infty|\langle x_0,e_n\rangle|^2=\lVert x_0\rVert^2 $$ and $\langle x_0,e_n\rangle =\Phi(e_n)$. In general, we cannot hope for a better $p$, that is, $p\lt 2$. Indeed, ...


2

It may be helpful to notice that if function $f$ is lipschitz and $f(0)=0$ then $f\circ u$ is sobolev with $\partial_i (f\circ u)(x)=f'(u)\partial_i u$. Hence, you could use $$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases} $$ as your truncation function. Of course $f$ satisfies all conditions I wrote above and of ...


2

the fact that $|fg_n|^p$ converges to $|fg|^p$ pointwise Where did you get this fact from? Convergence in $L^p$ does not imply convergence almost everywhere. However, the proof can be fixed. Convergence in $L^p$ does imply convergence almost everywhere for a subsequence $g_{n_k}$. So, for this subsequence your argument works and gives ...


2

From $L^p$ to $\omega$ If $f\in L^p$ with $1<p\le \infty$, then by Hölder's inequality. $$\int_E |f| \le \|f\|_p \|\chi_E\|_q = \|f\|_p(\mu(E))^{1/q} $$ where as usual, $p^{-1}+q^{-1}=1$. So, in this case $f$ has the modulus of integrability of the form $\omega(\delta)=C\delta^{1-\frac{1}{p}}$. This bound is not perfectly tight for $p<\infty$: ...


0

Hint: Compute for every p the limit $$\lim \frac{n^{1/n}-1}{n^{-p}}$$ using L'hospital's rule.


1

Hint: By Hölder's inequality, $$\begin{align*} \int_X |u_v \phi| \, dx &= \int_X |u_v| \cdot |1_{(0,1/v)}\phi| \, dx \leq \|u_v\|_p \cdot \|\phi 1_{(0,1/v)}\|_{p'}.\end{align*}$$ Now use monotone convergence in order to deduce that $$\int_X |u_v \phi| \, dx < \varepsilon$$ for $v$ sufficiently large.


2

You are on the right track. By Egoroff's, it is enough to show that for an arbitrary small set $A$ we have $$\int_A |f_n|^p dm \leq \epsilon.$$ By Holders, for $1/q+1/q'=1$, we get $$ \int_A |f_n|^p \cdot 1 \leq \left( \int_A |f_n|^{pq} dm \right)^{1/q} \left( \int_A 1^{q'} dm \right)^{1/q'}.$$ Choose $pq=2$ so that $1/q'=1-1/q=1-p/2=(2-p)/2$ or ...


0

The derivative of an $L^\infty$ function belongs to $W^{-1,\infty}$, a Sobolev space of negative order. This is a pretty abstract space, which is best understood as the space of distributional derivatives of bounded functions... that is, a tautology. The elements of $W^{-1,\infty}$ are linear functionals on $W^{1,\infty}_0$; indeed, if $g\in L^\infty$ ...


1

This is not a correct proof. You want to show that $\int_E |g|^q < \infty$, but then you are using this in the proof. So nothing has been shown. To write a correct proof, you will have to approximate $g$, as you wrote in your last comment. E.g.set $E_k = E \cap B_k(0)$ (so these sets now have finite measure) and $$ g_k(x) = \begin{cases} g(x) \quad (x ...


0

$\mu(\Omega\setminus \Omega_n)$ need not go to zero, consider $\Omega=(0,\infty)$ and $\Omega_n=(0,n)$. Use instead the Dominated convergence theorem, applied to the sequence $|f\chi_{\Omega_n}|^p\leq |f\chi_\Omega|^p\in L^1(\Omega)$.


1

There is a standard argument for problems like this: If you have a (compactly supported, continuous) function $g \in L^1((0,\infty))$, consider the convolution (calculated w.r.t. the Haar measure on $(0,\infty)$) $$ (f \ast g)(x)= \int_0^\infty f(x/y) g(y) \,dy/y. $$ The convolution is defined for every $x$ because of $f \in L^\infty$. Now an easy ...


0

What are the extreme points of $D = \{z \in \mathbb{C} : |z| \leq 1\}$? If $f \in C([0,1],\mathbb{C})$ has $f(x)$ an extreme point in $D$, and $f(x) = tg(x) + (1-t)h(x)$, what can we say about $g(x)$ and $h(x)$? Now what if $f(x)$ is an extreme point in $D$ for every $x \in [0,1]$?


1

The proof that $T$ is an isometry typically consists of two parts: Show that $\|Tx\|\le \|x\|$ for all $x$ Show that $\|Tx\|\ge \|x\|$ for all $x$ For the classical function and sequence spaces, 1) may involve Hölder's inequality, sometimes in its trivial $(1,\infty)$-form $\left|\sum a_nb_n\right|\le \sup_n|b_n| \sum_n |a_n|$. Step 2) then ...


0

I will answer, even if the title is inadequate. The question has been made clear in the comments. Let $$ n^{1/n}=1+h_n,\quad h_n>0,\quad \lim_{n\to\infty} h_n=0. $$ Then $$ n=(1+h_n)^n=\sum_{k=0}^n\binom{n}{k}h_n^k. $$ Take $k\in\mathbb{N}$. If $n>2k$ then $$ n\ge\binom{n}{k}h_n^k\implies h_n\le C_kn^{-(k-1)/k}, $$ where $C_k$ is a constant depending ...


4

$p'$ is the conjugate exponent of $p$. Let $(p-\epsilon)'$ be the conjugate exponent of $p-\epsilon$. Then $$ (p-\epsilon)'=p'+\frac{\epsilon}{(p-1-\epsilon)(p-1)}=p'+\delta,\quad\delta>0. $$ Then, by Hölder's inequality: $$ ...


0

I am interpreting your statement "decays like a polynomial" to mean something a bit different than what you wrote ($|x|^c$ goes to infinity with $x$ when $c > 0$, but I don't think you just meant that $g$ blows up at infinity), instead that there exists $M$ such that for $|x| \geq M$ $g(x) \leq C|x|^c$ for some $C, c$. Does that capture your intention? ...


4

This just uses the definition of weak derivative. The trick is to notice that if $u,\phi$ are both in $L^2(\mathbb R)$ then $$ \int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx = \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx \tag{$\ast$}.$$ If (2) holds, you can use the dominated convergence theorem and ($\ast$) to see that $$ ...


1

If the function $f$ is defined on the real line and vanishes at infinity, then combining the fundamental theorem of calculus with Hölder's inequality yields $$|f(x)|^2\leqslant 2\max\left\{\left(\int_0^{\infty}|f(t)|^p\mathrm dt\right)^{1/p}\left(\int_0^{\infty}|f'(t)|^q\mathrm dt\right)^{1/q}; \left(\int_{-\infty}^0|f(t)|^p\mathrm dt\right)^{1/p}\left( ...


1

Anything can happen. Take $f_m:=c_m\chi_{(0,a_m)} $ where $a_m\to 0$ and the unit interval is endowed with the Lebesgue measure. If $c_m=C a_m^{-1/m} $, the limit is $C$; if $c_m=m a_m^{-1/m}$, the limit is infinite; if $c_m=(-1)^ma_m^{-1/m}$, the limit does not exist.


4

Instead of just giving an example, I'll show a typical path to it. $L^p$ norms are invariant under rearrangement of the function: informally, we can move its values around, for example sorting them in decreasing order. So, it does not seem too restrictive to focus on decreasing functions. If $f$is decreasing, then $\int_{E} f\, dm\leq \int_{0}^{ m(E)} ...


1

First think about $\mathbb{R}$. In the context of Riemann integral, improper integrals are usually studied distinguishing between two cases: the case where the interval of integration is unbounded and the case where the interval of integration is bounded but the function is unbounded on that interval. Exploiting this idea you can build the examples that will ...


4

The Dominated Convergence Theorem works in $L^p$ spaces, too: if the dominating function is $L^p$ (with $1\le p < \infty$) then the convergence is in $L^p$ sense. To wit, assume that $$|f_n| \le g \in L^p,$$ with $p>1$ (the case $p=1$ being the DCT). Then the pointwise limit $f$ of $f_n$ (which exists by assumption) is dominated by $g$ and hence $f\in ...


5

Since $\Omega$ is bounded, we have a continuous inclusion $$j \colon L^q(\Omega) \hookrightarrow L^p(\Omega).$$ Now we have the general fact that a continuous linear map $T\colon X \to Y$ between two normed spaces (it holds more generally for Hausdorff locally convex spaces) is also continuous if we endow both spaces with their respective weak topology. ...


0

If you already know $(L^p)^*=L^q$, you can consider the isometric embedding $J:\ell^p\to L^p(0,\infty)$ that sends basis vectors $e_k$ to $\chi_{[k-1,k]}$. Every linear functional on $\ell^p$ can be viewed as a functional on $\operatorname{ran}J$, and from there extended (Hahn-Banach) to all of $L^p$. This gives a representation of the functional as some ...


1

This is Lebesgue dominated convergence theorem for the functions $f_\lambda:x\mapsto\lambda^p\,\mathbf 1_{|f(x)|\gt\lambda}$ such that $f_\lambda\to0$ pointwisely when $\lambda\to0$ and the domination $|f_\lambda|\leqslant g$ with $g=|f|^{1/p}$.


0

Take a simple function $g$ such that $\lVert f-g\rVert_p\lt\varepsilon$. Then $$\lambda^pF(\lambda)\leqslant \lambda^pm\{|f-g|>\lambda/2\}+\lambda^pm\{|g|>\lambda/2\}\leqslant 2^p\varepsilon+2^p\chi_{(0,\lVert g\rVert_\infty]}(\lambda).$$


1

You use the density of smooth compactly supported functions in $L^1(\mathbb{R})$. Namely, if $f\in L^1(\mathbb{R})$. Then there is a sequence of smooth compactly supported functions $f_n$ such that $\|f-f_n\|_{L^1(\mathbb{R})}\to0$ as $n\to\infty$. This is exactly what lines 6-7 of the solution say.


5

This is about the first thing proved in any text on the Fourier transform: $$ |\hat f(\xi)|=\Bigl|\int_{\mathbb{R}^n}e^{ix\xi}\,f(x)\,dx\Bigr|\le\int_{\mathbb{R}^n}|f(x)|\,dx=\|f\|_1. $$


3

If $(a_n)$ is in $l^2$ then $\sum |a_n|^2 < \infty$. Take $\epsilon >0$ and find an $n_0$ in $\mathbb{N}$ such that $\sum_{n\geq n_0} |a_n|^2 < \epsilon$. Then take the sequence $(b_n)$ given by $b_n = a_n$ if $n < n_0$ and $b_n=0$ otherwise. Is clear that $(b_n)$ is in $l^1$ because it is finitely supported. Observe that $||(a_n) - (b_n) ...


3

Let $\chi_{(a,b)} $be a characteristic function of an open interval $(a,b) .$ Take $f=\chi_{\left(0,\frac{1}{2}\right)} , g =\chi_{\left(0,\frac{1}{4}\right)} .$ Then $$||f||=||g|| =\frac{1}{2} ||f+g|| =1$$ but $$f\neq g.$$


0

Put $g_n = \chi[n,n+1) \in L^P(\mathbb R)$ for $n \in \mathbb N,$ i.e. $g_n$ is the characteristic function of the interval $[n,n+1).$ Then we have $\|g_n\|_p = 1.$ Now, define $f_{2n} = g_{2n}$ and $f_{2n-1} = 0$ for $n \in \mathbb N,$ i.e. $f_n$ is alternatingly $0$ or $g_n$. Note that $\|g_a-g_b\|_p \geq c > 0$ for $a\neq b$ with an appropriate $c = ...


0

You've misstated the definition of $L^p$. You should have $$ \|f_0\|^p = \left( \int_{\mathbb R^n} f_0^p \, dx \right)^{1/p}.$$ Where $dx$ represents integration with respect to Lebesgue measure. One way to evaluate the integral is to integrate $f_0^p$ over concentric rings centered at $0$. Given $0 < t < 1$ you can write $$ \{x : |x| < 1 \} = ...


5

The standard example for showing that $L^{\infty}(\Bbb{R})\not\subseteq L^p(\Bbb{R})$ would be $f(x) = 1$ for all $x\in\Bbb{R}$. An example in the other way would be the function $$f(x) = \begin{cases} n & x\in[n,n+\frac{1}{n^3}] \\ 0 & \text{otherwise}\end{cases}$$ here, $n$ is a natural number. The areas of each rectangular piece of $f$ are ...


0

Another counterexample even if $K \in L^1(\Omega \times \Omega)$. Consider $K(x,y) = \frac{1}{\sqrt{x}} \frac{1}{\sqrt{y}}$. And $\Omega \times \Omega = (0,1) \times (0,1)$. Put $\varphi \equiv 1$ in $(0,1)$. It is clear that $\varphi \in L^2(0,1)$. Then $$ \int_0^1 \left( \int_0^1 \frac{1}{\sqrt{x}} \frac{1}{\sqrt{y}} dy\right)^2 dx = \int_0^1 ...


1

It is true in general: Let $X_1$, $X_2$ be Banach spaces. And $f_1\in X_1^*$, $f_2\in X_2^*$. If $\Phi : X_1 \times X_2 \to \mathbb F$ is defined by $$ \Phi(x_1, x_2) = f_1(x_1) + f_2(x_2),$$ then $$||\Phi|| = \sup_{||x_1||+ ||x_2|| = 1} |f_1(x_1)| + |f_2(x_2)|$$ If you set $x_i = 0$, then you see that $||\Phi|| \geq \max\{ ||f_1||, ||f_2||\}$. On the ...


0

Let $U=\{x: f(x)\le 2\}$. This is a convex set. On the complement of $U$ we have $|f|^2> 4$, hence $|U^c|\le 1/4$ (writing $|\cdot|$ for the Lebesgue measure). This gives $|U|\ge 3/4$. Suppose $f(x_0)=-M$ for some $x_0$, where $M>2$. The set $V = \frac12x_0+\frac12U$ is contained in $[0,1]^d$ and has measure at least $3\cdot 2^{-d-2}$. By convexity, ...


0

Actually you can get, for a fixed $\epsilon>0$, that for any $u\in W^{m,p}(\Omega)$ $$ \|{u}\|_{W^{m,p}}\leq \epsilon\|{D^\alpha u}\|_{L^p}+C_{\epsilon}\|{u}\|_{L^q}, $$ where $|{\alpha}|=m$. Usually we just take $q=p$ but if domain is good enough and by embedding we could extend $q$ to where embedding would go. This theorem states that the extreme term ...


4

Take $$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$ for $n = 1, 2, \dots$ and for some $a_n$. You'll get $$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$ $$\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$ Then choose the sequance $a_n$ so that the first sum is convergent, while the second one ...


0

How about this one? $$f(x)=\frac1{x(1-\ln(x))^2}$$ Then $$\int f(x)\,dx=\frac1{1-\ln x},$$ and it quickly follows that $f\in L^1$. If $p>1$, pick $r$ with $p>r>1$ and write $$f(x)^p=\frac1{x^r}\cdot\frac1{x^{p-r}(1-\ln(x))^{2p}},$$ note that the first factor is non-integrable and the second factor is bounded below (it is continuous and positive on ...


0

$$f = \sum_n \frac{1}{2^n \cdot \int_0^1 x^{-1+1/n} \, dx} \cdot x^{-1 + 1/n}$$ EDIT: The idea is that $x^{-1 + 1/n}$ is "barely integrable", but not integrable if taken to a power $p > \frac{-1}{-1 + 1/n} \to 1$. Hence, we sum all these functions in such a way that the result is still integrable (use monotone convergence). But because of $f \geq C_n ...


1

To elaborate on the comment, let $N > 1/p$ be a natural number and $f$ a Schwartz function. Then $|f(x)| \leq C (1+|x|)^{-nN}$ for some $C>0$ and all $x$. Hence, $$ \int |f(x)|^p \, dx \leq C \int (1+|x|)^{-n \cdot Np} <\infty, $$ where the last integral is finite because of $-n \cdot Np < -n$.


1

Hint: Consider the indicator functions of intervals.


0

We have $\|\partial^\beta u\|_{L^p(U)} \leq \|u\|_{W^{k-1,p}(U)}$ Also, by the theorem, $W^{k,p}(U) \rightarrow W^{k-1,p}(U)$ is compact. Now, suppose there is no $C_\epsilon$. Then we get a sequence $u_n \in W^{k,p}(U)$ (normalize so these are all norm 1) violating it with $n$ in place of $C_\epsilon$. That is, $\|u_n\|_{W^{k-1,p}} \geq \epsilon + n ...


4

Suppose $r>0$. Notice that $x^{-r}$ is in $L^p$ for each $p<1/r$ and is not in $L^p$ for each $p \geq 1/r$. So take a sequence $\{ r_n \}_{n=1}^\infty$ such that $1/r_n$ decreases to $1$. For example, $r_n = 1-2^{-n}$. Now the idea is to choose $c_n>0$ so small that $$f(x)=\sum_{n=1}^\infty c_n x^{-r_n}$$ is in $L^1$. For instance we can pick ...


1

divide the interval [0,1] to N subintervals and take the function $F_N$ as below: in odd subintervals equal 1 in even subintervals equal 0



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