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## Hot answers tagged lp-spaces

3

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.

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 $L^1$.

3

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

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 $x,y\in [0,1]$ $$f_n(x)-f_n(y)=\int_y^x f_n^{\prime}(t)dt$$

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

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 $||f\ast g||_{\infty}\leq ||f||_1||g||_{\infty}$.

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 $$T_Nf(x):=\int_{|\alpha |\leq N}\hat f(\alpha )e^{2i\pi x\alpha }\... 2 Let f\in C^\infty_c(\Omega). Then f is supported in a compact set K and |f| attains a maximum C in this K. Thus$$\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. 2 The p "norm" fails to satisfy the triangle inequality for p<1. 2 Use the dominated convergence theorem with \max(1,|f(x)|^2) as the dominating function. 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 ... 1 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. 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 ... 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 ... 1 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). 1 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

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).

1

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\... 1 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$\int_n^{n+1}|g|^q\to0$. 1 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 ... 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 $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 $$\|f\ast g\|_\infty \leq \|f\|_1\|g\|_\infty$$ 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; ... 1 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 ...

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