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3

Let $s$ integrable and $\varepsilon$ such that $s\leqslant|f|$ on $X$ and $\displaystyle\int_X|f|\leqslant\varepsilon+\int_Xs$. Then, for every measurable $E\subseteq X$, $|f|-s\geqslant0$ on $X\setminus E$ hence $\displaystyle\varepsilon\geqslant\int_X|f|-s=\int_E|f|-s+\int_{X\setminus E}|f|-s\geqslant\int_E|f|-s$ , which implies ...


0

The proposition is trivial if the function $f$ is bounded. So assume that $f_n(x) = n$ if $f(x) \leq n$ and $f_n(x) = 0$ otherwise. Then each $f_n$ is bounded and $f_n \to f$ pointwise so by the Monotone convergence theorem $\int_E f_n \to \int_E f$. So given $\epsilon > 0$ there exists an $N$ such that $\int_E f - \int_E f_N < \epsilon/2$. Choose ...


0

Summary of comments Fix $a$ and let $T(x)=\sum_{k=1}^\infty a_kx_k$. For $n=1,2,\dots$ define linear functionals $T_nx = \sum_{k=1}^n a_kx_k$. These are bounded on $\ell^2$ and satisfy $\sup_n |T_n(x)|<\infty$ for every $x$, because the series $\sum_{k=1}^\infty a_kx_k$ converges. By the uniform boundedness principle, also known as the ...


2

And another proof: Let $\epsilon>0$ and choose $a$ large enough so that $\|f 1_{[a,\infty)} \|_p < {\epsilon \over 2}$. Now choose $L \ge a$ large enough so that $|\frac{1}{x^{1-\frac1{p}}}\int_0^a f(t)\,dt | < {\epsilon \over 2}$ whenever $x \ge L$. Then we have $|\frac{1}{x^{1-\frac1{p}}}\int_0^x f(t)\,dt | = |\frac{1}{x^{1-\frac1{p}}}\int_0^a ...


4

Use an approximation argument: it is true when $f$ is a simple function (linear combination of characteristic function), and for any $s$ simple and each positive $x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$


0

When we consider integrability, estimates within a constant factor are enough. Then the sum of two positive numbers is as good as their maximum, because $$\max(a,b)\le a+b\le 2\max(a,b)$$ So, replace $x^2+y^a$ by $\max(x^2,y^a)$. Then we only need to integrate $$ \iint \chi_{\{x^2<y^a\}} y^{-a} \,dx\,dy + \iint \chi_{\{x^2>y^a\}} x^{-2} \,dy\,dx $$ ...


1

Because $L^{p}$ spaces expose the subtle nature of arguments. You have reflexive, non-reflexive, separable, non-separable, algebra, Hilbert, Banach, etc.. And, interpolation works between such spaces because of the exponent. They're good spaces for testing conjectures. They're the original spaces that firmly established the need to separate a space from its ...


2

For $1<p<\infty$, strict convexity of the norm implies that the only shortest path is the line segment between these points. For $p=1$, the length of a path $(x(t),y(t))$ is just the sum of the lengths of one-dimensional paths $x(t)$ and $y(t)$. Both of those must go from $0$ to $1$. They will have length $1$ if and only if the function is ...


2

Instead of $(0,T)\times\Omega$ work on $I=[0,1]$. Let \begin{align*} f_n(x)=\begin{cases} x^{-2} & x\in[1/n;1]\\ 0 & x\in [0;1/n] \end{cases} \end{align*} Then $|\{|f_n|>k\}|\leq |\{x^{-2}>k\}|=k^{-1/2}$ but $\lim_{n\to\infty}\int_0^1f_n(x)dx=\infty$.


2

(Partial answer) For your question that there does not exists a function $g$ increasing on $[0,1]$ such that for all $0\leq a<b\leq 1$ we have $$g(b)-g(a)\geq \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}}$$ you can argue as follows. First, changing $g$ to $g(x)-g(0)$ if necessary, we can suppose that $g(0)=0$. Then if $b>a=0$, we get $g(b)\geq 1$. Now ...


3

I don't see any use of any Hardy-Littlewood inequality here. You have a measure space $(Q,\delta^\alpha \,dx\,dt)$, which has finite total measure. I will denote the measure by $\mu$ for simplicity. The assumption (2.7) says that $u$ is in the weak $L^{\hat q}( d\mu)$ space. Then it's just a matter of interpolation to get that $u\in L^q( d\mu)$ for every ...


0

The following is not a complete answer. In the following let $f$ be some candidate for a counterexample. First observation: $f$ cannot be monotone. If it were, say, increasing, then we would have $$\int_0^x |f'| dx \leq f(x)-f(0)$$ by the Lebesgue decomposition theorem. Similarly $\int_0^x |f'(x)| dx \leq f(0) - f(x)$ if $f$ is decreasing. Second ...


2

If $p\gt 1$, we define the $L^{p,\infty}$ semi-norm by $$\lVert f\rVert_{p,\infty}^p:=\sup_{t\gt 0}t^p\lambda\{s, |f(s)|\gt t\}$$ (this is equivalent to a norm, namely, $\sup_{A,\lambda(A)\in (0,\infty)}\mu(A)^{1/p-1}\int_A|f|\mathrm d\lambda$). If we define $x:=k^{1/m}$ and if we use the inequality, we obtain $$x^{2m\frac{N+1}N}\lambda\{|u|\lt ...


1

Lemma. If $g\colon\mathbf R^n\to\mathbf R$ is continuous and $g(x)\leqslant 0$ for each $x\in A$, then $g(x)\leqslant 0$ for each $x\in \overline A$. We use this with $g(x):=|f(x)|-K$ and $A:=\Omega\setminus N$. To show the lemma, take $x$ in the closure of $A$ and $(x_n)_{n\geqslant 1}$ a sequence of elements of $A$ converging to $x$. Then ...


1

Well the biggest problem is that it is a classical result of analysis/topology that the space of continuous functions is closed under the $L^\infty$ topology i.e. the topology of a.e. uniform convergence (in the case of continuous functions we take a continuous representative, and a convergent sequence in uniformly Cauchy, thus converges uniformly to a ...


1

Taking for simplicity $\Omega$ compact, if $C^\infty_c(\Omega)$ were to be dense in $(L^\infty,\|\cdot\|_\infty)$ then having that $C^\infty_c(\Omega)$ is separable because every differentiable function can be approximated by rational polynomials, so would have to be $L^\infty(\Omega)$.


1

The compactly supported part is most important. Try to $L^\infty$-approximate the constant function $g \equiv 1$ by a compactly supported function $f$. There will be a set of positive measure where $f$ is zero, so $\| f - g \|_\infty$ will always be at least $1$ no matter what $f$ we pick.


0

Alternatively you can use Hölder's Inequality to find that $$ \lVert x \rVert_r \le \lVert x \rVert_p \tag{1} $$ and then you can proceed by letting $\epsilon > 0$ and putting $\delta = \epsilon$ so that $$ \lVert x - y \rVert_p < \delta \implies \lVert i(x) - i(y) \rVert_p < \epsilon \implies \lVert i(x) - i(y) \rVert_r < \epsilon $$ Hint for ...


0

We have $\|x\|_r \le \|x\|_p$ (from Jensen's inequality, using $\lambda[0,1] = 1$). Let $i:L^p[0,1] \to L^r[0,1]$ be the injection. Then $\|i(x)-i(y)\|_r = \|x-y\|_r\le \|x-y\|_p$, hence it is Lipschitz continuous. This nesting is true more broadly for $L^p(X,\mu)$, where $\mu X < \infty$ (except the Lipschitz constant changes).


1

Consider $h_n = f \cdot \chi_{\Bbb{R}^n \setminus B_{1/n}(\gamma_0)}$ and use dominated convergence. This implies that $h_n \to f$ in $L^p$ and each $h_n$ is constant on the ball $B_{1/n}(\gamma_0)$ with radius $1/n$ around $\gamma_0$. EDIT: Ok, to get a solution to the precise formulation of your problem, take $$ h(x) = (f(x) - f(\gamma_0)) \cdot ...


1

The duality $\ell^p(X)^*=\ell^q(X^*)$ for $1<p<\infty$ holds for every Banach space. Indeed, $c_{00}(X)$, the space of finitely supported sequences, is dense in $l^p(X)$. Therefore, every linear functional on $l^p(X)$ is determined by its values on sequences with one nonzero element. This identifies such a functional with an $X^*$-valued sequence ...


2

The sum must extend over all integers, by the way, otherwise, it would be finite (namely $0$) for a constant function whose value is less than $1$ in modulus. Then, with $B_n = \{x : 2^n < \lvert f(x)\rvert \leqslant 2^{n+1}\}$, we have $$A_n = \{ x : 2^n < \lvert f(x)\rvert\} = \bigcup_{k=n}^\infty B_k,$$ and the union is disjoint. Thus ...


3

Hint: Holder's inequality with $f$ and $g(x)=1$.


1

Suppose that $L^p$ is closed in $L^1$: in order to prove the existence of $C$, we only need to show that $L^1\subset L^p$, then we use the closed graph theorem. Let $f\in L^1$; define $f_n:=f\chi_{\{|f|\leqslant n\}}$. Then the sequence $(f_n)_{n\geqslant 1}$ is an element of $L^p$, which converges in the $L^1$ norm to $f$. By closeness, $f$ belongs to ...


3

If $u\in\mathbb L^p$ for some $p>1$, then take $u_n:=u\chi_{\{|u|\leqslant n\}}$. If $u$ does not belong to any $\mathbb L^p$ space for any $p>1$, then it is not possible: if $\lVert u_n-u\rVert_1\to 0$ and $(u_n)_n$ is bounded in $\mathbb L^p$, then extract a subsequence $(u_{n_k})_{k\geqslant 1}$ which converges almost everywhere to $u$. Then using ...


3

Hint: $\Omega=(0,1)$, $f(x)=\log x$.


0

Since my attempt to close the question as a duplicate failed, I'll post an answer: the statement follows from a general theorem on the continuity of Nemytskii operator, which is stated and proved here.


3

Suppose $u_n \stackrel*\rightharpoonup u$ in some $X^*$. Given $\epsilon > 0$ choose some $x\in X$ with $\|x\| = 1$ and $|u(x)| \ge \|u\|-\epsilon$. We have $$ \lim |u_n(x)| = |u(x)| \ge \|u\| - \epsilon $$ and on the other hand $$ \lim |u_n(x)| \le \liminf \|u_n\|\|x\| = \liminf \|u_n\| $$ So $$ \|u\| - \epsilon \le \liminf\|u_n\| $$ for each ...


2

This is the answer I've unravelled: Let $z \in (\ell^p)^{\ast\ast}$. I want to prove that exists $x \in \ell^p$ such that $\langle z,f\rangle=\langle f,x \rangle$ for every $f \in (\ell^p)^\ast$. I know that there are the isomorphisms: $j_p: \ell^q \rightarrow (\ell^p)^\ast$ and $j_q: \ell^p \rightarrow (\ell^q)^\ast$ Now, fix $z \in ...


0

It is not true that $$x=\sum_{n=1}^{\infty} x_ne_n$$ for $x\in\ell^{\infty}$, because convergence does not necessarily occur in the $\ell^{\infty}$-norm. Indeed, if $x=(1,1,1,\ldots)$, then, for any $N\in\mathbb N$, $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(\underbrace{0,\ldots,0}_{N\text{ times}},1,1,1,\ldots)\|_{\infty}=1,$$ so $x$ cannot be the ...


1

Note that "$x = \sum_n x_n e_n$" does not hold in norm if $x \in \ell^\infty \setminus c_0$, neither does it weakly, as for $x \not\in c_0$ there is some $x^*\in (\ell^\infty)^*$ with $x^*|_{c_0} = 0$ and $x^*(x) = 1$. Then $$ x^*\left(\sum_{n=1}^N x_ne_n\right) = 0 \not\to 1 = x^*(x) $$ It holds only weakly$^*$, as given $y \in \ell^1$, we have $xy\in ...


0

The sum you give is not norm convergent, unless $x\in c_0$. (It is weakly$\!^*$ convergent, but that is beside the point.) You have here a Schauder basis for $c_0$, a (comparatively) tiny subspace of $l^\infty$.


1

No. If $f_\epsilon(x)>f(x)$, then $-f_\epsilon(x)<-f(x)$, so for any $f$, either $f$ or $-f$ is a counterexample. The convergence is decreasing when $f$ is subharmonic. This is because $\frac{d}{d\epsilon}f_\epsilon$ is the convolution of $f$ against a radially symmetric function that is negative on small radii and positive for large ones. The ...


1

So you have the canonical maps: $$\begin{cases} f_q: \ell^p\to (\ell^q)^* \\ f_p: (\ell^q)^*\to (\ell^p)^{**}\end{cases}$$ which are the isomorphisms between $\ell^p$ and $(\ell^q)^*$ and $\ell^q$ and $(\ell^p)^*$ respectively. Then just write $f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$.


1

Here's another explanation of the failure of inequality $$\tag{1} \|f\ast g\|_{L^p(\mathbb{R}^N)}\le C \|f\|_{L^p(\mathbb{R}^N)}\|g\|_{L^p(\mathbb{R}^N)},\qquad f, g\in \mathcal{S}$$ for $p>1$. (The OP is about $N=1$ but there is no added difficulty in considering the general case). It is a routine application of the so-called scaling argument. Assume ...


2

The inclusion stated in the title follows from the fact that $\mathcal{S}\subset L^1$. But the inequality would involve the $L^1$ norm of $f$, not its $L^p$ norm. (Namely, Young's inequality for convolution.) The point is, smoothness is irrelevant to $L^p$ norm estimates of this sort. To see why you can't have $\|f\|_{L^p}$, consider $f=\chi_{[0,M]}$ ...


0

If $1<p<2$, it holds. Note that $f \in C(R)$ implies $f \in L^{\infty}$. Since $f$ is in $L^p \cap L^\infty$, we get $f \in L^2$. Thus we get the desired result. For $f \in L^p$ with $ 2<p<\infty$, Fourier transform of $f$ is not defined unless $f \in L^q$ for some $1 \leq q \leq 2$.


0

This is false. Consider, for example, $f_n=n^{3/4}\chi_{(0,1/n)}$ on $D=[0,1]$. Then $f_n\to 0$ in $L^1$, but the sequence does not converge weakly in $L^2$ (it would be bounded in norm if it did). The problem with your argument is that you are implicitly assuming that $\|f_n\|_2$ is bounded; otherwise, there's no obvious way to carry out the approximation ...


3

By Jensen's inequality $\int |f|^2 \log |f|=\int |f|^2 \cdot \frac{1}{p-2}\log |f|^{p-2} = \frac{1}{p-2}\cdot\int |f|^2\log |f|^{p-2} \leq \frac{1}{p-2}\log (\int |f|^{p-2}\cdot |f|^2) = \frac{1}{p-2}\cdot \frac{p}{2}\log (\int|f|^p)^\frac{2}{p}=\frac{1}{p-2}\cdot \frac{p}{2} \log ||f||_p^2$ because $\frac{1}{p-2}=\frac{n-2}{4}, ...


1

The denseness of $C_c^\infty(\def\R{\mathbb R}\R)$ in $L^p(\R)$ does the trick. Just use Hölder's inequality. For $f \in L^p(\R)$, choose $f_n \in C^\infty_c(\R)$ with $\def\norm#1{\left\|#1\right\|}\norm{f_n -f}_p \to 0$. Then $$ \left|\left<f_n - f, g\right>\right| \le \norm{f_n - f}_p \norm g_q \to 0$$ Addendum: We will show $\sup B = \alpha$. ...


1

The inclusion $$ (C([0,1]), || \cdot ||_{\infty}) \longrightarrow (L^2([0,1]),|| \cdot ||_2) $$ is continuous since for all $f \in C([0,1])$ $$|| f ||_2^2 = \int_0^1 f^2 \leq \int_0^1 ||f||_{\infty}^2 = ||f||_{\infty}^2 $$ If you restrict the inclusion to $M$ $$i: (M, || \cdot ||_{\infty}) \longrightarrow (M,|| \cdot ||_2) $$ is continuous and ...


0

Define $\iota\colon (M,\lVert \cdot\rVert_2)\to (C[0,1],\lVert\cdot\rVert_\infty)$ by $\iota(f)=f$. We have to show that $\iota$ is continuous. Since $M$ and $C[0,1]$ endowed with their respective norms are complete, we have to show that if $\lVert f_n-f\rVert_2\to 0$ and $\lVert f_n-g\rVert_\infty=0$ then $f=g$. To see this, notice that $f_n\to g$ a.e. ...


1

Yes it is true because $C^\infty_c(\mathbb{R})$ is dense in $L^p(\mathbb{R})$ for any $1\leq p<\infty$ in the strong topology and is dense in $L^\infty$ in the weak-$\star$ topology. Edit: $L^\infty(\mathbb{R})$ is the dual of $L^1(\mathbb{R})$ and $C^\infty_c(\mathbb{R})$ is dense in $L^\infty(\mathbb{R})$. For any $f\in L^1(\mathbb{R})$ then the scalar ...


1

Adding to martini's answer: From $$ \|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta $$ one finds using Young's inequality in the form $$ ab \le \theta a^{\frac1\theta} + (1-\theta)b^{\frac1{1-\theta}} $$ the estimate $$ \|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta \le (1-\theta)\|f\|_{p_0} + \theta\|f\|_{p_1}. $$ Using the definition of $p$ it ...


1

There is some convexity, don't know if this helps you, but: Let $p_0, p_1 \in [1,\infty]$, $\theta \in [0,1]$, $\frac 1p = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$. Then for $f \in L^{p_0}\cap L^{p_1}$, by Hölder \begin{align*}\def\norm#1#2{\left\|#1\right\|_{#2}}\def\abs#1{\left|#1\right|} \norm fp &= \norm{\abs{f}^{1-\theta}\abs{f}^\theta}p \\ ...


2

One can forget $p(s)$, $r(s)$ and the rest and simply try to show that, for every $a\lt b$, $$\|u\|_a\leqslant\|u\|_b.$$ To wit, considering $v=|u|^a$ and $p=b/a\gt1$, note that Hölder inequality yields $$ \int |u|^a=\int v\leqslant\left(\int v^p\right)^{1/p}=\left(\int |u|^b\right)^{a/b}, $$ that is, $$ \left(\int |u|^a\right)^{1/a}\leqslant\left(\int ...


5

No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let $r_k$ be the $k$th Rademacher function on $[0,1]$, that is $r_k = \operatorname{sign}\sin ( 2^k \pi x) $. Then $2^p r_k \rightharpoonup 2^{p-1}\mathbf {1}$ in $L^1$, where $\mathbf{1}$ is the constant function equal to $1$. On the other hand, ...


0

Both assumptions imply convergence in the sense of distributions: that is, for every smooth compactly supported function $\varphi$ we have $$ \int \varphi u_n\to \int \varphi u,\qquad \int \varphi u_n\to \int \varphi v \tag1$$ So, $$\int \varphi u = \int \varphi v\tag2$$ for every such $\varphi$. This means exactly that $u=v$ in the sense of distributions. ...


1

Looking at non-negative unsigned Lebesgue integrable functions on $\mathbb{R}$, step function: taking finitely many values on finitely many intervals. simple function: taking finitely many values on finitely many Lebesgue measurable sets. True: Both step functions and simple functions are dense in $L^1$. They can approximate any Lebesgue integrable ...


1

I expect you mean to ask if you can find a monotonically increasing sequence of step functions approximating the given function from below? The answer is a resounding no. For a simple example, consider the characteristic function of a fat Cantor set.



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