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1

Ashamed of myself for not realizing the problem was so elementary, I'll register my solution here, in case it helps someone. We have that for all $a,b \in \Bbb R$: $$\sqrt{a^2 + b^2} \leq \sqrt{|a|^2 + 2|ab| + |b|^2} = \sqrt{(|a|+|b|)^2} = ||a|+|b|| = |a|+|b|$$ and: $$(|a|-|b|)^2 \geq 0 \implies a^2 - 2|ab| + b² \geq 0 \implies |ab| \leq 2|ab| \leq a^2 + ...


0

Suppose $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ is a Schwartz class function satisfying the support of $\varphi$ is contained in $\{ \omega : 2^{-1} \leq |\omega| \leq 2 \}$ $\varphi(\omega)>0$ for $2^{-1} <|\omega| <2$ $\sum_{k\in\mathbb{Z} } \varphi(2^{-k}\omega) =1$ for $\omega \neq 0$ Define \begin{equation*} \widehat{f}_N (\omega) ...


0

The only mistake I see is where you claim that $|f|\leq|f_{n_1}|+|g|$ implies $|f|^p\leq|f_{n_1}|^p+|g|^p$ , which is not true ($1\leq1/2+1/2$ does not imply $1^2\leq (1/2)^2+(1/2)^2$). What you have to use is that the sum of two functions in $ L^p $ is again in $ L^p $.


2

We have the relationship $\lVert f\rVert_p\leqslant \lVert f\rVert_q$ for each function $f$. Since the embedding is linear, it follows that it is continuous. The image contains $\mathbb L^\infty$, which is dense. It can be written as a countable union of closed sets with empty interior.


1

If one knows the value at one point and has an inequality, one can actually derive an inequality for the derivative as follows: Let us assume for simplicity that $f : [0,\infty) \to \Bbb{R}$ fulfils $f(x) \leq 0$ for all $x$ and $f(0) = 0$. Then (if $f$ is (one-sidedly) differentiable at $0$), $$ f'(0) = \lim_{h \downarrow 0} \frac{f(h) - f(0)}{h} \leq 0, ...


0

One of the problems here is that each element of your spaces is really an equivalence class of a.e. equal functions and not a genuine function. This is especially a problem in the case $u:(0,T)\to L^2$, because one has trouble selecting for each $t$ some representative function $x \mapsto u(t)(x)$. One way to circumvent this is the following. Observe that ...


2

If $1<p<q<\infty$, and $f\in L^q(0,1)$, then $$ \int_0^1 \lvert\, f(x)\rvert^p\,dx=\int_0^1 \lvert\, f(x)\rvert^p\cdot 1\,dx \le \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}\left(\int_0^1 1^r\,dx\right)^r= \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}, $$ where $$ \frac{p}{q}+\frac{1}{r}=1\quad\text{or}\quad r=\frac{q}{q-p}. $$ Hence ...


0

Hint: apply the Holder inequality with the parameters $\frac qp > 1$ and $$ \frac{\frac qp}{1-\frac qp} $$


1

Certainly not. If $g$ is in the image of $A$, it has the property that $x\mapsto g(x)(1-x)$ is constant, which not all elements of $L^\infty(\mathbb R,L^1([0,1],\mathbb R))$ have


3

Observe that for $p<\infty$ $$ l_p\cap k = l_p \cap k_0 = l_p. $$ Hence these spaces are complete with respect to the $l_p$ norm. To see these inclusions: If $x\in l^p$, then $\sum_{n=1}^\infty |x_n|^p$ converges, hence $\lim_{n\to\infty}x_n=0$ and $x\in k_0$. These spaces are not complete with respect to the $l_\infty$ norm: Take the sequence $x_n$ ...


0

Let $f\in L^p(\mathbb{R}^n)$. The canonical distribution corresponding to $f$ is defined by $d_f:\mathcal{S}(\mathbb{R}^n)\rightarrow \mathbb{C}$, $g\mapsto \int_{\mathbb{R}^n} fg$. Now check that $d_f$ is an element of $\mathcal{S}^\prime(\mathbb{R}^n)$. It is clear that it is a linear map. So it remains to check continuity. Namely, we have to check that ...


1

If $f\in L^p(\mathbb R^n)$ and $\varphi\in\mathscr S(\mathbb R^n)$, then $$ \ell(\varphi)=\int_{\mathbb R^n}f\,\varphi, $$ is definable, since $f\varphi\in L^1(\mathbb R^n)$, as $\varphi\in L^q(\mathbb R^n)$, for all $p$, and $$ \lvert \ell(\varphi)\rvert\le\|\varphi\|_{p'}\|\,f\|_p, $$ where $\frac{1}{p}+\frac{1}{p'}=1$. All the other properties are ...


1

Answers. a. The essential range is not necessarily achieved by averages. For examples if $f(x)=x$, $x\in [0,1]$, then the values 0 and 1 are not achieved by $A_f$. b. Indeed the range of $A_f$ is not necessarily convex. Let $\mu=\delta_0+\delta_1$, then the only values it achieves are $f(0)$, $f(1)$ and $\frac{1}{2}\big(f(0)+f(1)\big)$. It is convex in ...


2

The $L^p$-spaces with $p=1,2,\infty$ get the most attention due to their special properties: $L^2$ is Hilbert, $L^\infty$ and $L^1$ are not reflexive, $L^\infty$ is not separable, $L^1$ has no predual, and so on. Viewed as plain Banach spaces, maybe the spaces $L^p$, $p\not \in\{1,2,\infty\}$ are just boring spaces with no special and interesting properties. ...


-2

I would say that there are very few applications of the cases where $p \ne 1,2, \infty$. One example: in $\mathbb R^2$ and $\mathbb R^3$, the balls corresponding to the $l^p$ norms are so-called super ellipses and superquadrics. These are things that are shaped somewhat like regular ellipses, but with corners that become "sharper" as $p$ increases. These ...


0

You can show that its complement is open. Let $X = (c_o)^c = \{\{a_n\}_1^\infty: \{a_n\}_1^\infty \ \text{does not converge to 0}\}$. That means that if $x \in X$ then there exists $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there exists $n > N$ such that $|x_n - 0| = |x_n| > \epsilon$. Now we'll find an $\epsilon$-neighborhood about x ...


0

Note that the space $c_0 \subset c \subset \ell_{\infty}$ is a, not dense, hyperplane (kernel) of the operator $T \colon c \to \mathbb{R}$ $$T(x) = \lim\limits_{n \to \infty} x_n$$


1

Hint: Let $(u_n)$ a sequence of element of $c_0$ such taht $u_n \to u$ in $\ell^{\infty}$ Let $\varepsilon > 0$, then : $$\exists N \in \mathbb N, n \geq N \Rightarrow \|u_n-u\|_{\infty} < \varepsilon $$ In particular : $ \|u_N-u\|_{\infty} < \varepsilon/2 $ Since $u_N \in c_0$ we have: $$\exists N_1 \in \Bbb N, \forall p \geq N_1 ...


0

As pointed out in the comment, the set $T$ is not closed, but $$\left\{ \left\{ x_i \right\} \in {\ell ^\infty }:\left| x_i \right| \leqslant \mu_i,\mathop \lim\limits_{i \to \infty } \mu _i = 0 \right\}$$ is. We can show that it is precompact by fixing $\varepsilon$. We then take $N$ such that $|\mu_i|\lt\varepsilon$ if $i\geqslant N+1$ and we use ...


1

1) By Holder's Inequality it is immediate to show that $L^p(\Omega) \subset L^q(\Omega)$ for $p \ge q$ if you consider the usual Lebesgue measure on a set of finite measure. 2) Note that $\sum x^p$ converges if $\sum x^q$ converges provided $p \ge q$. This means that $\ell^q(\mathbb{N}) \subset \ell^p(\mathbb{N})$, which we can rewrite as ...


0

This is true if $\mathbb R$ is replaced by a closed interval $[a,b]$. To see that this is not true, take $$ f_n(x)=\left\{\begin{array}{clc} 0 & \text{if} & x\le -2n, \\ \frac{x+2n}{n} & \text{if} & -2n\le x\le -n, \\ 1 & \text{if} & -n\le x \le n, \\ \frac{2n-x}{n} & \text{if} & n\le x\le 2n, \\ 0 & \text{if} & x ...


1

The metric on X is induced by the norm, so that if $x = \{x_i\}$ and $y = \{y_i\}$ are elements of X, then $d(x,y) = \sup_{i \in \mathbb{N}} |x_i - y_i|$. Now I claim that $X_{++}$ is not an open set. To show a set in a metric space is open, I have to show that for any element in that set, I can find an open ball around that element entirely contained in ...


2

Try to write $k \mapsto (z- 4 \pi^2 |k|^2)^{-1}$ as the Fourier transform of a function $h\in L^1$ (try a multiple of $x \mapsto e^{-\alpha |x|}$, with $\alpha$ a square root of $-z$ - notice the condition on $z$ is what ensures you can pick a square root with strictly positive real part which is needed to get it to be in $L^1$), then recall $$\mathcal F h ...



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