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Hot answers tagged lp-spaces

5

The standard example for showing that $L^{\infty}(\Bbb{R})\not\subseteq L^p(\Bbb{R})$ would be $f(x) = 1$ for all $x\in\Bbb{R}$. An example in the other way would be the function $$f(x) = \begin{cases} n & x\in[n,n+\frac{1}{n^3}] \\ 0 & \text{otherwise}\end{cases}$$ here, $n$ is a natural number. The areas of each rectangular piece of $f$ are ...

5

Since $\Omega$ is bounded, we have a continuous inclusion $$j \colon L^q(\Omega) \hookrightarrow L^p(\Omega).$$ Now we have the general fact that a continuous linear map $T\colon X \to Y$ between two normed spaces (it holds more generally for Hausdorff locally convex spaces) is also continuous if we endow both spaces with their respective weak topology. ...

5

This is about the first thing proved in any text on the Fourier transform: $$|\hat f(\xi)|=\Bigl|\int_{\mathbb{R}^n}e^{ix\xi}\,f(x)\,dx\Bigr|\le\int_{\mathbb{R}^n}|f(x)|\,dx=\|f\|_1.$$

4

The Dominated Convergence Theorem works in $L^p$ spaces, too: if the dominating function is $L^p$ (with $1\le p < \infty$) then the convergence is in $L^p$ sense. To wit, assume that $$|f_n| \le g \in L^p,$$ with $p>1$ (the case $p=1$ being the DCT). Then the pointwise limit $f$ of $f_n$ (which exists by assumption) is dominated by $g$ and hence $f\in ... 4 This just uses the definition of weak derivative. The trick is to notice that if$u,\phi$are both in$L^2(\mathbb R)$then $$\int_{\mathbb R} \frac{u(x+h) - u(x)}{h} \phi(x) \, dx = \int_{\mathbb R} u(x) \frac{\phi(x-h) - \phi(x)}{h} \, dx \tag{\ast}.$$ If (2) holds, you can use the dominated convergence theorem and ($\ast$) to see that $$... 4 p' is the conjugate exponent of p. Let (p-\epsilon)' be the conjugate exponent of p-\epsilon. Then$$ (p-\epsilon)'=p'+\frac{\epsilon}{(p-1-\epsilon)(p-1)}=p'+\delta,\quad\delta>0. $$Then, by Hölder's inequality:$$ ... 4 Suppose$r>0$. Notice that$x^{-r}$is in$L^p$for each$p<1/r$and is not in$L^p$for each$p \geq 1/r$. So take a sequence$\{ r_n \}_{n=1}^\infty$such that$1/r_n$decreases to$1$. For example,$r_n = 1-2^{-n}$. Now the idea is to choose$c_n>0$so small that $$f(x)=\sum_{n=1}^\infty c_n x^{-r_n}$$ is in$L^1$. For instance we can pick ... 4 Instead of just giving an example, I'll show a typical path to it.$L^p$norms are invariant under rearrangement of the function: informally, we can move its values around, for example sorting them in decreasing order. So, it does not seem too restrictive to focus on decreasing functions. If$f$is decreasing, then$\int_{E} f\, dm\leq \int_{0}^{ m(E)} ...

4

Take $$f(x) = a_n \quad\text{ if }\quad x\in\left(\frac{1}{2^n},\frac{1}{2^{n-1}}\right]$$ for $n = 1, 2, \dots$ and for some $a_n$. You'll get $$\int_0^1 f(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$$ $$\int_0^1 f^p(x)\, dx = \sum_{n=1}^{\infty}\frac{a_n^p}{2^n}$$ Then choose the sequance $a_n$ so that the first sum is convergent, while the second one ...

3

Let $\chi_{(a,b)}$be a characteristic function of an open interval $(a,b) .$ Take $f=\chi_{\left(0,\frac{1}{2}\right)} , g =\chi_{\left(0,\frac{1}{4}\right)} .$ Then $$||f||=||g|| =\frac{1}{2} ||f+g|| =1$$ but $$f\neq g.$$

3

If $(a_n)$ is in $l^2$ then $\sum |a_n|^2 < \infty$. Take $\epsilon >0$ and find an $n_0$ in $\mathbb{N}$ such that $\sum_{n\geq n_0} |a_n|^2 < \epsilon$. Then take the sequence $(b_n)$ given by $b_n = a_n$ if $n < n_0$ and $b_n=0$ otherwise. Is clear that $(b_n)$ is in $l^1$ because it is finitely supported. Observe that $||(a_n) - (b_n) ... 2 From$L^p$to$\omega$If$f\in L^p$with$1<p\le \infty$, then by Hölder's inequality. $$\int_E |f| \le \|f\|_p \|\chi_E\|_q = \|f\|_p(\mu(E))^{1/q}$$ where as usual,$p^{-1}+q^{-1}=1$. So, in this case$f$has the modulus of integrability of the form$\omega(\delta)=C\delta^{1-\frac{1}{p}}$. This bound is not perfectly tight for$p<\infty$: ... 2 First,$A\subset l^2$is the collection of sequences that only take finitely many values, their entries are all zero after some point (you said constant, but the constant must be zero), and$A$is dense in$l^2$. Now let us show$B$is dense in$A$(which would mean$B$is also dense in$l^2$), let$a\in A$and$\epsilon$be given, we show there exists a ... 2 It may be helpful to notice that if function$f$is lipschitz and$f(0)=0$then$f\circ u$is sobolev with$\partial_i (f\circ u)(x)=f'(u)\partial_i u$. Hence, you could use $$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases}$$ as your truncation function. Of course$f$satisfies all conditions I wrote above and of ... 2 You are on the right track. By Egoroff's, it is enough to show that for an arbitrary small set$A$we have $$\int_A |f_n|^p dm \leq \epsilon.$$ By Holders, for$1/q+1/q'=1$, we get $$\int_A |f_n|^p \cdot 1 \leq \left( \int_A |f_n|^{pq} dm \right)^{1/q} \left( \int_A 1^{q'} dm \right)^{1/q'}.$$ Choose$pq=2$so that$1/q'=1-1/q=1-p/2=(2-p)/2$or ... 2 the fact that$|fg_n|^p$converges to$|fg|^p$pointwise Where did you get this fact from? Convergence in$L^p$does not imply convergence almost everywhere. However, the proof can be fixed. Convergence in$L^p$does imply convergence almost everywhere for a subsequence$g_{n_k}$. So, for this subsequence your argument works and gives ... 1 Anything can happen. Take$f_m:=c_m\chi_{(0,a_m)} $where$a_m\to 0$and the unit interval is endowed with the Lebesgue measure. If$c_m=C a_m^{-1/m} $, the limit is$C$; if$c_m=m a_m^{-1/m}$, the limit is infinite; if$c_m=(-1)^ma_m^{-1/m}$, the limit does not exist. 1 Hint: Consider the indicator functions of intervals. 1$p=2$does the job. To see this, represent the bounded linear form$\Phi$as$\Phi(x)=\langle x_0,x \rangle$. Bessel's equality gives the wanted convergence. Indeed, we have $$\sum_{n=1} ^\infty|\langle x_0,e_n\rangle|^2=\lVert x_0\rVert^2$$ and$\langle x_0,e_n\rangle =\Phi(e_n)$. In general, we cannot hope for a better$p$, that is,$p\lt 2$. Indeed, ... 1 It is true in general: Let$X_1$,$X_2$be Banach spaces. And$f_1\in X_1^*$,$f_2\in X_2^*$. If$\Phi : X_1 \times X_2 \to \mathbb F$is defined by $$\Phi(x_1, x_2) = f_1(x_1) + f_2(x_2),$$ then $$||\Phi|| = \sup_{||x_1||+ ||x_2|| = 1} |f_1(x_1)| + |f_2(x_2)|$$ If you set$x_i = 0$, then you see that$||\Phi|| \geq \max\{ ||f_1||, ||f_2||\}. On the ... 1 Hint: By Hölder's inequality, \begin{align*} \int_X |u_v \phi| \, dx &= \int_X |u_v| \cdot |1_{(0,1/v)}\phi| \, dx \leq \|u_v\|_p \cdot \|\phi 1_{(0,1/v)}\|_{p'}.\end{align*} Now use monotone convergence in order to deduce that $$\int_X |u_v \phi| \, dx < \varepsilon$$ forv$sufficiently large. 1 This is Lebesgue dominated convergence theorem for the functions$f_\lambda:x\mapsto\lambda^p\,\mathbf 1_{|f(x)|\gt\lambda}$such that$f_\lambda\to0$pointwisely when$\lambda\to0$and the domination$|f_\lambda|\leqslant g$with$g=|f|^{1/p}$. 1 You use the density of smooth compactly supported functions in$L^1(\mathbb{R})$. Namely, if$f\in L^1(\mathbb{R})$. Then there is a sequence of smooth compactly supported functions$f_n$such that$\|f-f_n\|_{L^1(\mathbb{R})}\to0$as$n\to\infty$. This is exactly what lines 6-7 of the solution say. 1 Alternatively and directly: Assume by contradiction that$(e_n)_{n}$does not converge weakly to zero. Then there exists a$\epsilon >0$and a functional$f\in l_\infty^\ast$s.t.$|f(e_n)|\geq \epsilon$for infinitely many$n\in\mathbb{N}$. By passing to that subsequence, we have that$|f(e_{n_k})|\geq \epsilon$for all$k\in\mathbb{N}$. Let$\lambda_k ...

1

First think about $\mathbb{R}$. In the context of Riemann integral, improper integrals are usually studied distinguishing between two cases: the case where the interval of integration is unbounded and the case where the interval of integration is bounded but the function is unbounded on that interval. Exploiting this idea you can build the examples that will ...

1

The proof that $T$ is an isometry typically consists of two parts: Show that $\|Tx\|\le \|x\|$ for all $x$ Show that $\|Tx\|\ge \|x\|$ for all $x$ For the classical function and sequence spaces, 1) may involve Hölder's inequality, sometimes in its trivial $(1,\infty)$-form $\left|\sum a_nb_n\right|\le \sup_n|b_n| \sum_n |a_n|$. Step 2) then ...

1

The span $S$ of the functions $\{ \chi_{[0,x]} : 0 \le x \le 1\}$ is a dense linear subspace of the Hilbert space $L^{2}[0,1]$. To see that the span is dense, it is enough to show that $(f,\chi_{[0,x]})=0$ for all $x$ and some $f \in L^{2}[0,1]$ implies $f=0$. By the Lebesgue differentiation theorem, the following holds a.e.: $$... 1 divide the interval [0,1] to N subintervals and take the function F_N as below: in odd subintervals equal 1 in even subintervals equal 0 1 If the function f is defined on the real line and vanishes at infinity, then combining the fundamental theorem of calculus with Hölder's inequality yields$$|f(x)|^2\leqslant 2\max\left\{\left(\int_0^{\infty}|f(t)|^p\mathrm dt\right)^{1/p}\left(\int_0^{\infty}|f'(t)|^q\mathrm dt\right)^{1/q}; \left(\int_{-\infty}^0|f(t)|^p\mathrm dt\right)^{1/p}\left( ...

1

This is not a correct proof. You want to show that $\int_E |g|^q < \infty$, but then you are using this in the proof. So nothing has been shown. To write a correct proof, you will have to approximate $g$, as you wrote in your last comment. E.g.set $E_k = E \cap B_k(0)$ (so these sets now have finite measure) and  g_k(x) = \begin{cases} g(x) \quad (x ...

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