New answers tagged

0

Suppose that $f\in L^p$ for some $p\in(2,\infty)$. (If $p=\infty$, simply take $f_1=0$ and $f_2=f$.) The case in which $\|f\|_p=0$ is trivial (take $f_1=f_2=0$), so suppose that $\|f\|_p>0$. Let $$A\equiv\{x\in X\,|\,|f(x)|>2\|f\|_p\}$$ and define $f_1\equiv f\times\mathsf I_A$ and $f_2\equiv f\times \mathsf I_{A^\mathsf c}$. Clearly, $f=f_1+f_2$. ...


1

Actually you don't even need the factor of $2$ there. Indeed, take $f_1=f \cdot 1_{\{x:\;|f(x)|> \| f\|_p\}}$ and $f_2=f \cdot 1_{\{x:\;|f(x)| \leq \|f\|_p\}}$. Clearly, $\|f_2\|_{\infty} \leq \|f\|_p$ by definition. On the other hand, if $a,b$ are positive real numbers with $a>b$ then $a/b>1$ so (since $p>2$) we see that $(a/b)^p>(a/b)^2$ ...


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Here is a proof of the inequality: first Hoelder, then estimate $L^1$ against $L^2$, Fubini, estimate the inner integral, done: $$ \frac1{h^2}\int_0^{t_1}\int_\Omega \left|\int_t^{t+h} f(s) ds\right|^2dx \ dt \le\frac1{h^2}\int_0^{t_1}\int_\Omega \left(\int_t^{t+h} |f(s)| ds\right)^2dx \ dt\\ \le\frac1h\int_0^{t_1}\int_\Omega \int_t^{t+h} |f(s)|^2 ds\ dx \ ...


1

Here's a proof why $l^p(\mathbb N)$ is not locally convex, this is just for simplicity, it can be easily generalized. If it would be locally convex, then the unit ball $B_1(0)$ would contain a convex neighborhood U of $0$. Then there must be $\delta>0$ with $B_{2\delta}(0)\subset U$, hence also $\mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0)$. Let ...


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This is standard notation: if $p \in \mathcal D ' (\Omega)$, then $\nabla p \in \mathcal D ' (\Omega, \Bbb R^d)$ is a vector-valued distribution (the terminology is misleading, because it suggests that a vector-valued distribution has vectors as values, which it does not; the term is quite natural, though, in that a vector-valued distribution is a ...


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This is not true. Take for example $p=1$ and the function $f_\epsilon(x)=\frac{1}{x}$ for $x>\epsilon$ and 0 else. Then $f_\epsilon$ is in $L^1(\epsilon,T)$ with norm equal to $ln(\epsilon)-ln(T)$. Bug the norm diverges as $\epsilon$ goes to 0. Hence $\frac{1}{x}$ is not in $L^1(0,T)$.


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There may be other reasons, but simple in order to integrate something which depends on $f_h$, $f_h$ has to be well defined. So if you integrate from $0$ to $t_1$, you need $t_1+h \le T$, or, equivalently, $t_1 \le T-h < T$ (since you have to assume $h>0$ to make sense of this).


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Use the dominated convergence theorem with $\max(1,|f(x)|^2)$ as the dominating function.


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Hint: Suppose $\theta = 1/2$ just to scratch around a bit. For $h>0$ we have $$|\int_a^{a+h} f\,\,|^p \le \frac{1}{2}\cdot h^{p-1}\int_a^{a+h} |f|^p \implies |\frac{1}{h} \int_a^{a+h} f|^p \le \frac{1}{2}\cdot \frac{1}{h}\int_a^{a+h} |f|^p.$$ In the last inequality, let $h\to 0^+$ and apply the Lebesgue differentiation theorem.


2

The $p$ "norm" fails to satisfy the triangle inequality for $p<1$.


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We endow the unit interval with Borel $\sigma$-algebra and Lebesgue measure. Denote for $n\geqslant 1$ and $0\leqslant k\leqslant 2^n-1$ the interval $I_{n,k}:=\left[k/2^n,(k+1)2^{-n}\right)$. If $N$ is such that $2^n\leqslant N\lt 2^{n+1}$ for some $n\geqslant 1$, define $f_N(x):=n\mathbf 1_{\left(I_{n,N-2^n}\right)}(x)$. In this way, $\lVert f_N\...


0

HINT If we let $$ g = \limsup_{n \rightarrow \infty} f_n,$$ then there is a subsequence $\{f_{n_k} \}$ of $\{ f_n \}$ such that $f_{n_k} \rightarrow g$ as $k \rightarrow \infty$.


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As a comment said, we use the Cauchy-Schwarz inequality: $$\|f\|_{L^1([-a,a])}=\|f\cdot1\|_{L^1([-a,a])}\leq\|f\|_{L^2([-a,a])}\|1\|_{L^2([-a,a])}=\|f\|_{L^2([-a,a])}(2a)^{1/2}<\infty$$ by the given fact that $\|f\|_{L^2([-a,a])}<\infty$. But perhaps a more fundamental way to prove it is Jensen's inequality with the function $\varphi(x)=x^2$.


1

Hints, assuming $1<p<\infty$ (the answer is slightly different for the endpoint cases): First, if $f_n\to0$ weakly in $L^p$ then $||f_n||_p$ is bounded (by the Uniform Boundedness Principle). Second, if $g\in L^q$ then $\sum\int_n^{n+1}|g|^q<\infty$, hence $\int_n^{n+1}|g|^q\to0$.


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The case $\alpha < 0$ is ok. The next calculation is also ok, but not the conclusion (as pointed out by Arctic chair). However, if you replace your $g$ by $g_n$ and observe that $g_n$ has constant norm in $L^q$, you find that the norm of $f_n$ has to be unbounded in $L^p$ if $\alpha > 0$ (note that $n^\alpha \equiv 1$ for $\alpha = 0$). For $\alpha = ...


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Not sure this is the fastest and most elegant solution, but this fact is rather a standard fact from distribution theory. $\textbf{First case}:$ If $\Omega$ is bounded than it is obvious (as you wrote), since set has finite Lebesgue measure. $\textbf{Second case}:$ Set $\Omega$ is open and unbounded. This is not trivial fact (but a good excercise) to show ...


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Let $f\in C^\infty_c(\Omega)$. Then $f$ is supported in a compact set $K$ and $|f|$ attains a maximum $C$ in this $K$. Thus $$\int_{\Omega} |f|^p dx = \int_K |f|^p dx \le \int_K C^p dx = \text{Vol}(K) C^p.$$ Thus $f\in L^p$ for all $p$. Indeed $C^\infty_c(\Omega)$ is dense in $L^p$ for all $1\le p <\infty$.


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I'm assuming that $C^\infty_c(\Omega)$ is the space of $C^\infty$ with compact support contained in $\Omega$. If that's the case, then you don't care whether or not $\Omega$ has finite measure, since $$ \int_\Omega |u|^pdx\leq |spt(u)|\cdot\max |u|^p<\infty. $$ where $spt(u)$ is the support of $u$, that is defined as $$ spt(u)=\overline{\{x\in \Omega :...


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Well yeah it has compact support, hence outside of a compact set $K$ is zero. It's continuos on a bounded, closed set hence its integral must be smaller than $M\cdot m(K)$, where $M$ is the maximum of the function on $K$ and $m$ is the lebesgue measure


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Note that $L^p(\Omega)$ is not a set of functions, but a set of equivalence classes of functions; specifically, we define an equivalence relation by saying for functions $f, g : \Omega \rightarrow \mathbb{R}$, we have $f \sim g$ if and only if the set $\{ \omega \in \Omega | f(\omega) \not= g(\omega) \}$ has measure 0 (it can be checked that this does indeed ...


1

It doesn't matter - the two versions of the definition give isometrically isomorphic spaces. Allowing functions to be undefined on a set of measure zero can be convenient, for example allowing us to refer to $f(x)=|x|^{-1/2}$ as an element of $L^1([-1,1])$ without having to define $f(0)$. Or allowing us to define $f=\lim f_n$ when the limit only exists ...


0

For the first question, since $|G'(s)|\le M$ for all $s\in\mathbb{R}$, it follows that $|G(u)-G(u_n)|\le M|u-u_n|$, and hence $\|G\circ u - G\circ u_n\|_{L^p(\Omega)}\le M\|u-u_n\|_{L^p(\Omega)}\rightarrow 0$. For the second question, notice that $$ (G'\circ u)\frac{\partial u}{\partial x_i} - (G'\circ u_n)\frac{\partial u_n}{\partial x_i} = (G'\circ u - G'\...


1

We assume that $p>2$. Note that \begin{align*} |u(t)|^p &\le K_1 \int_0^t \left(1+|u(s)|^2\right)|u(s)|^{p-2} ds + K_2 \\ &=K_1 \int_0^t |u(s)|^{p-2} ds + K_1 \int_0^t |u(s)|^{p} ds + K_2. \end{align*} Let $\alpha = (p-2)/p$, and \begin{align*} v(t) = K_1 \int_0^t |u(s)|^{p-2} ds + K_1 \int_0^t |u(s)|^{p} ds + K_2. \end{align*} Then, \begin{align*...


1

You can approximate any function in the Lebesgue space arbitrarily well by a differentiable function. Taking a converging sequence of such approximating functions you can show that $ Vf_n $ is Cauchy in the Sobolev space, hence has a limit in the Sobolev space. Since $ V $ is continuous as operator into the Lebesgue space the Sobolev limit equals $ Vf $


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Your work is indeed correct although you do not need to consider the cases $p=1$ and $p>1$ separately. In both cases the scenarios are identical.


2

Yes; apply Holder's inequality (you can also use Jensen if you like): \begin{align*} \|f\|_{L^p}=\|f\cdot1\|_{L^p}\leq\|f\|_{L^q}\|1\|_{1/(1/p-1/q)}=\|f\|_{L^q}\mu(X)^{1/(1/p-1/q)}\leq\|f\|_{L^q} \end{align*} since we assumed $\mu(X)<1$, and $1/(1/p-1/q)\geq0$ from $p\leq q$. Edit: Please note I used a slightly generalized version of the Holder ...


2

A bunch of things: The perspective that $L^p$ spaces have been very successful is probably skewed heavily by your, excuse me for being blunt, limited exposure to research literature. The reason that $L^p$ spaces appear frequently in textbooks is because that they are simple to define and thus serve a great purpose pedagogically. That said, $L^p$ spaces (...


2

There is no "obvious proof". The proof is essentially based on the fact that Hilbert transform is $L^p-$bounded (i.e. that $$\|Hf\|_{L^p(\mathbb R)}\leq C\|f\|_{L^p(\mathbb R)}$$ for a suitable constant $C$. The proof of this is not that easy. But this boundness will allow you to show that $$T_Nf(x):=\int_{|\alpha |\leq N}\hat f(\alpha )e^{2i\pi x\alpha }\...


0

As has been mentioned in a comment your question is very broad. One possible answer is to use the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$, it consists of the linear functional continuous with respect to $\mathcal{S}(\mathbb{R}^n)$ which is the Schwartz space. In other words it can be shown that the ...


1

By Hölder's inequality, you get $$ |f\ast g(x)| \leq \int_{\mathbb{R}^{n}} |f(y)g(x-y)|\;{\rm d}y = \|f g(x-\cdot)\|_1\leq \|g\|_\infty \|f\|_1 $$ since $|g(x-y)|\leq \|g\|_\infty$ for almost every $x\in\mathbb{R}^{n}$ since $\|g(x-\cdot)\|_\infty=\|g\|_\infty$. This implies $$ \|f\ast g\|_\infty \leq \|f\|_1\|g\|_\infty $$


2

If $f\in L^1$ and $g\in L^{\infty}$, then $$ |(f\ast g)(x)|=\Big|\int_{\mathbb{R}}f(x-y)g(y)\;dy\Big|\leq \int_{\mathbb{R}}|f(x-y)g(y)|\;dy\leq ||g||_{\infty}\int_{\mathbb{R}}|f(y)|\;dy=||f||_1||g||_{\infty}$$ using the translation-invariance of Lebesgue measure. Therefore $||f\ast g||_{\infty}\leq ||f||_1||g||_{\infty}$.


1

Take a smooth function which is supported on $[0, e^{-n}]$ and with height $n$. One way to do this is to take the characteristic function of $[e^{-n}/4, 3e^{-n}/4]$ and mollify it, then multiply by $n$. Then these functions converge to zero in not only $L^2$, but also every $L^p$ with $p < \infty$. They're each compactly supported, smooth, and bounded; ...


3

Consider $f_n = n\chi_{[0,1/n^3]}$ on $[0,1].$


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This map is not onto. The image consists of mean zero functions in the sense that $\int_{\mathbb{R}} Sf(x) dx=0$. Roughly speaking this is because $$\int_\mathbb{R} Sf(x)dx=\int_\mathbb{R}\int_0^1 (f(x)-f(x+y)) dy dx = \int_0^1 \left( \int_\mathbb{R} f(x) dx - \underbrace{\int_\mathbb{R} f(x+y) dx}_{=\int_\mathbb{R} f(x) dx} \right) dy = 0. $$ Of course ...


2

Hint: As $f_n$ is continuously differentiable, you have $$f_n(x)=\int_0^x f_n^{\prime}(t)dt+f_n(0)$$ and hence, for $x,y\in [0,1]$ $$f_n(x)-f_n(y)=\int_y^x f_n^{\prime}(t)dt$$


3

Let $r > 0$. Then $$ \int_{0}^{\infty}e^{-rx}e^{-isx}dx=\frac{1}{r+is}. $$ The function $f(x)=e^{-rx}\chi_{[0,\infty)}(x)$ is in $L^1$, but $\hat{f}(s)=\frac{1}{\sqrt{2\pi}(r+is)}$ is not in $L^1$.



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