Tag Info

Hot answers tagged

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, ...


4

Rewrite the integral as follows $$\int_0^1 |f(x)|^p|g(x)|dx = \int_0^1 \int_0^{|f(x)|} pt^{p-1}|g(x)| dt dx$$ Switching the order of integration we obtain $$\int_0^\infty \int_{|f(x)|>t} pt^{p-1}|g(x)| dx dt = \int_0^\infty pt^{p-1}\int_{|f(x)|>t} |g(x)| dx dt$$ Now note that $\int_{|f(x)|>t} |g(x)| dx \leq \min(\frac{3}{t^2},|g|_{L^1})$ Hence the ...


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 $x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$


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 ...


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}, ...


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 ...


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


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 ...


3

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


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 ...


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 $f=\chi_{[0,M]}$ ...


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 ...


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

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 ...


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 ...


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

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 ...


2

To prove the first inequality it suffices to prove that $(x+y)^{p}\le x^{p}+y^{p}$ for $x,y\ge0$ and $p\in(0,1)$. Notice that we can use this as follows: $\vert x\rvert^{p}\le\lvert x-y\rvert^{p}+\lvert y\rvert^{p}$ then reversing the roles of $x$ and $y$. Notice that this Lemma can be proven by reducing to the case when $y=1$ then looking at ...


2

For the first problem, use that any $L^2$-convergent sequence has a subsequence that converges almost everywhere. Alternatively, you can use that any finite-dimensional subspace of any normed vector space is closed. This essentially follows (in your case), because $$\Bbb{R} \rightarrow \Bbb{R}_+, x \mapsto \Vert x \cdot \chi_{[0,1]} \Vert_{L^2}$$ gives a ...


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 ...


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 ...


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

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 ...


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 ...


1

For the constants: prove that the subspace is complete. In an Hilbert space, this is equivalent to being closed, and in this case it is easier. For the null integrals subspace: show that $\int f_n\to \int f$.


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.


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

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

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.



Only top voted, non community-wiki answers of a minimum length are eligible