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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, $f=f_1+f_2$. ...

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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: $$\frac1{h^2}\int_0^{t_1}\int_\Omega \left|\int_t^{t+h} f(s) ds\right|^2dx \ dt \le\frac1{h^2}\int_0^{t_1}\int_\Omega \left(\int_t^{t+h} |f(s)| ds\right)^2dx \ dt\\ \le\frac1h\int_0^{t_1}\int_\Omega \int_t^{t+h} |f(s)|^2 ds\ dx \ ... 1 Here's a proof why l^p(\mathbb N) is not locally convex, this is just for simplicity, it can be easily generalized. If it would be locally convex, then the unit ball B_1(0) would contain a convex neighborhood U of 0. Then there must be \delta>0 with B_{2\delta}(0)\subset U, hence also \mathrm{conv}(B_{2\delta}(0))\subset U\subset B_1(0). Let ... 0 This is standard notation: if p \in \mathcal D ' (\Omega), then \nabla p \in \mathcal D ' (\Omega, \Bbb R^d) is a vector-valued distribution (the terminology is misleading, because it suggests that a vector-valued distribution has vectors as values, which it does not; the term is quite natural, though, in that a vector-valued distribution is a ... 1 This is not true. Take for example p=1 and the function f_\epsilon(x)=\frac{1}{x} for x>\epsilon and 0 else. Then f_\epsilon is in L^1(\epsilon,T) with norm equal to ln(\epsilon)-ln(T). Bug the norm diverges as \epsilon goes to 0. Hence \frac{1}{x} is not in L^1(0,T). 1 There may be other reasons, but simple in order to integrate something which depends on f_h, f_h has to be well defined. So if you integrate from 0 to t_1, you need t_1+h \le T, or, equivalently, t_1 \le T-h < T (since you have to assume h>0 to make sense of this). 2 Use the dominated convergence theorem with \max(1,|f(x)|^2) as the dominating function. 3 Hint: Suppose \theta = 1/2 just to scratch around a bit. For h>0 we have$$|\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.$$In the last inequality, let h\to 0^+ and apply the Lebesgue differentiation theorem. 2 The p "norm" fails to satisfy the triangle inequality for p<1. 1 We endow the unit interval with Borel \sigma-algebra and Lebesgue measure. Denote for n\geqslant 1 and 0\leqslant k\leqslant 2^n-1 the interval I_{n,k}:=\left[k/2^n,(k+1)2^{-n}\right). If N is such that 2^n\leqslant N\lt 2^{n+1} for some n\geqslant 1, define f_N(x):=n\mathbf 1_{\left(I_{n,N-2^n}\right)}(x). In this way, \lVert f_N\... 0 HINT If we let$$ g = \limsup_{n \rightarrow \infty} f_n,$$then there is a subsequence \{f_{n_k} \} of \{ f_n \} such that f_{n_k} \rightarrow g as k \rightarrow \infty. 1 As a comment said, we use the Cauchy-Schwarz inequality:$$\|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$$by the given fact that \|f\|_{L^2([-a,a])}<\infty. But perhaps a more fundamental way to prove it is Jensen's inequality with the function \varphi(x)=x^2. 1 Hints, assuming 1<p<\infty (the answer is slightly different for the endpoint cases): First, if f_n\to0 weakly in L^p then ||f_n||_p is bounded (by the Uniform Boundedness Principle). Second, if g\in L^q then \sum\int_n^{n+1}|g|^q<\infty, hence \int_n^{n+1}|g|^q\to0. 0 The case \alpha < 0 is ok. The next calculation is also ok, but not the conclusion (as pointed out by Arctic chair). However, if you replace your g by g_n and observe that g_n has constant norm in L^q, you find that the norm of f_n has to be unbounded in L^p if \alpha > 0 (note that n^\alpha \equiv 1 for \alpha = 0). For \alpha = ... -2 Not sure this is the fastest and most elegant solution, but this fact is rather a standard fact from distribution theory. \textbf{First case}: If \Omega is bounded than it is obvious (as you wrote), since set has finite Lebesgue measure. \textbf{Second case}: Set \Omega is open and unbounded. This is not trivial fact (but a good excercise) to show ... 2 Let f\in C^\infty_c(\Omega). Then f is supported in a compact set K and |f| attains a maximum C in this K. Thus$$\int_{\Omega} |f|^p dx = \int_K |f|^p dx \le \int_K C^p dx = \text{Vol}(K) C^p.$$Thus f\in L^p for all p. Indeed C^\infty_c(\Omega) is dense in L^p for all 1\le p <\infty. 0 I'm assuming that C^\infty_c(\Omega) is the space of C^\infty with compact support contained in \Omega. If that's the case, then you don't care whether or not \Omega has finite measure, since$$ \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 ...

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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 ...

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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  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. 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 }\...

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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 ...

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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 $$\|f\ast g\|_\infty \leq \|f\|_1\|g\|_\infty$$

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

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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; ...

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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 ...

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

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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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