# Tag Info

0

Suppose $f\in L_{w}^{p}$. Fix $\lambda>0$ and define $f_{0,\lambda}:=f\chi_{\left|f\right|\geq\lambda}$ and $f_{1,\lambda}:=f-f_{0,\lambda}$. I claim that $f_{0,\lambda}\in L_{w}^{p_{0}}$ and $f_{1,\lambda}\in L_{w}^{p_{1}}$. Indeed, \begin{align*} ...

1

Elements of $L^q$ are precisely those classes of measurable functions with $\| f \| _q < \infty$. Now if you have such an $f \in L^q$, due to the inequality you have found you will also have $\| f \| _p \le C \| f \| _q < \infty$, so $f \in L^p$, therefore $L^q \subset L^p$.

2

Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy$$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz.$$ This is just the convolution operator $$f\mapsto \phi*f$$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$ Now one of the fundamental inequalities (not hard to prove) about ...

2

$Tf=\phi*f$, where $\phi(t)=e^{-t}\chi_{(0,\infty)}(t)$. (I had $\phi$ backwards in the first version; noticed that guest's $\phi$ was different, then noticed his was right.) So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In ...

1

The counterexample doesn't hold true: Because $\sup_{x\in[0,1]}x^{-\frac{1}{2}}=\infty$ it follows $f\notin L^{\infty}([0,1])$, so the pairing $\langle f,f\rangle$ is not defined.For $f\in L^p,g\in L^q$, $p^{-1}+q^{-1}=1$ the dual pairing $\langle f,g\rangle$ is always finite, because of Hölder's inequality.

1

"Scalar product" is a misnomer. It's actually a duality, that is, a bilinear map $L^p\times L^q\to\mathbb F$.

0

For $1 < p < \infty$, the space $\ell^p$ is reflexive, so the weak and weak-star topologies on $B$ coincide, and thus $B$ is a compact Hausdorff space by the Banach-Alaoglu theorem. The polynomials in question form a subalgebra of $C(B)$ that separates points, and by the Stone-Weierstrass theorem this algebra is dense in $C(B)$. The same result fails ...

5

It's certainly true for $1<p<\infty$. In that case the weak topology on $B$ is the same as the weak* topology regarding $\ell^p$ as the dual of $\ell^{q}$. So $B$ is a compact Hausdorff space, and so Stone-Weierstrass implies that those polynomials are dense in $C(B)$. Note that you didn't specify what topology on $C(B)$ you're talking about when you ...

2

Since $f$ is $L_p$-integrable, $f$ is $L_1$-integrable. By Lusin's theorem, there exists a function $h$ which is continuous in $[0,1]$ such that $$\mu(\{x:|f(x)-h(x)|>\epsilon^{1/p}/(2\mu(\Omega))\})<\delta$$ Let $A=\{x:|f(x)-h(x)|>\epsilon^{1/p}/(2\mu(\Omega))\}$. Then $$\int_{\Omega} |f-h|^p\,d\mu = \int_{\Omega-A} |f-h|^p\,d\mu +\int_{A} ... 0 we prove \int |f|^p\,d\mu is bounded first. Since f_n \to f in measure, then there is a subsequence f_{n_k} converge a.e. to f. By Fatou's lemma$$ \int_{\Omega} |f|^p\,d\mu = \int_{\Omega} \lim\limits_{k \to \infty} |f_{n_k}|^p\,d\mu \leq \liminf\limits_{k \to \infty} \int_{\Omega} |f_{n_k}|^p \,d\mu\leqslant\left(\int_{\Omega} ...

4

By Holder's inequality, $$\int_{|f_n-f|>\alpha} |f_n - f| d\mu \leq \int_{|f_n-f|>\alpha} |f_n| d\mu + \int_{|f_n-f|>\alpha} |f| d \mu \leq$$ $$\mu(\{\omega\in\Omega:|f_n(\omega)-f(\omega)|>\alpha\})^{1/q} \left(||f_n||_p+||f||_p\right)$$ As $n\to\infty$, $\mu(\{\omega\in\Omega:|f_n(\omega)-f(\omega)|>\alpha\})\to 0$, and ...

1

I'm guessing that the question is to show that $\phi$ is locally integrable. This is very simple, doesn't use any fancy theorems. You have $$\phi=g*f,$$where $g(x)=1/|x|$. Now $g=g_1+g_2$, where $g_1\in L^1(\Bbb R^3)$ and $g_2\in L^\infty(\Bbb R^3)$. So $g_1*f\in L^1$, while $g_2*f$ is continuous (and bounded).

0

Let $f_n \to f$ in $L^p(\Omega)$. Then, we also have that $\|f_n\|_p \to \|f\|_p$. So "something similar" holds. As for the convergence of $\int f_n$ to $\int f$, this is generally not guaranteed by $L^p$ convergence, unless the measure of the underlying space is finite (like it is in your example). In that case we have $$\left| \int_{\Omega} f_n(x) dx - ... 0 Actually your equalities should be$${\chi _{[0,x]}}(x + t) = \left\{ {\begin{array}{*{20}{c}} {1,}&{x + t \in [0,x]}\\ {0,}&{else} \end{array}} \right. = \left\{ {\begin{array}{*{20}{c}} {1,}&{t \in [ - x,0]}\\ {0,}&{else} \end{array}} \right. = \chi_{[-x,0]}(t)$$, so that {\chi _{[0,x]}}(x) = \mu([-x,0]) . Now we have  ... 0 No, this doesn't exists. Consider the sequences b^{(i)}i such that b^{(i)}_n = 1 if i=n and 0 elsewhere Take an arbitrary N \in \Bbb N. Then as e^{(r)} are an approximation of the identity, there exists n_0 such that \forall n > n_0, \forall 1 \leq i \leq N, \| e^{(n)} b^{(i)} - b^{(i)} \| \leq \epsilon This imply that \forall i ... 6 Pretty sure it's false - maybe you should check with the guys who wrote the exam. It's going to be a counterexample in L^2([0,1]), with g=1. Say (I_n)_{n=1}^\infty is a sequence of disjoint intervals with |I_n|=2^{-n}. Define$$F_1=\chi_{I_1}$$and$$F_n=2^{n-1}\chi_{I_n}-2^{n-2}\chi_{I_{n-1}}\quad(n>1).$$Then \sum F_n=0 almost everywhere. ... 1 In case it's not clear, if you assume only u\in L^1 then no, you can't say anything nice. First: Of course in that case you really want to consider the essential supremum, not the sup: If you let u be the characteristic function of the rationals then u=0 almost everywhere although \sup_{|y|<h}|u(x+y)-u(x)|=1 for every x. That's a totally dumb ... 0 Take n such that j,k \geq n implies$$ \|f_j - f_k\|_\infty < 1\,. $$Now$$ \|f_j\|_\infty \leq \|f_j - f_n\|_\infty + \|f_n\|_\infty \leq 1 + \|f_n\|_\infty\,. $$We know that |f_j(x)| \leq \|f_j\|_\infty a.e. x and |f(x)| = \lim_{j \to \infty} |f_j(x)|. What can you say about this limit now? To show that \|f_n - f\|_\infty \to 0, use the ... 0 Hint: First prove Minkowski's Integral Inequality$$ || || f || _{L^1(P)} || _{L^r(Q)} \leq || || f ||_{L^r(Q)} || _{L^1(P)} $$using Holder's Inequality(writing F(y) = \int|f(x,y)| d \mathbb{P}(x) may make things more apparent for you). The result to your question follows by writing q/p=r and using Minkowski's integral inequality. 2 As mentioned in the comments, by the Hölder inequality (and because \mu(\Omega)=1) we have$$ \|f\|_1 = \int_{\Omega} |f(x)|\,d\mu(x) \leq \left(\int_{\Omega}|1|^2\,d\mu(x)\right)^{\frac12}\left(\int_{\Omega}|f(x)|^2\,d\mu\right)^{\frac12} = \|f\|_2. $$Then, if f_n \to f in L^2 we have 0\leq \|f_n-f\|_1 \leq \|f_n-f\|_2\to 0, thus f_n \to f in ... 2 You are almost right. Just use the Cauchy-Schwarz inequality$$\mu(\Omega)=\int_{\Omega} 1 \,d\mu=\int_{\Omega} \left|f^{1/2}\frac{1}{f^{1/2}}\right|\,d\mu\leqslant \left(\int_{\Omega}|f|\, d\mu\right)^{\frac{1}{2}}\left(\int_{\Omega}\frac{1}{|f|}\, d\mu\right)^{\frac{1}{2}} $$1 The usual way to estimate L^p norms using weak estimates is by using the Layer-Cake formula (see L^p-norm of a non-negative measurable function)$$ \int |f|^q \, d\mu = \int_0^\infty q \cdot \lambda^{q-1} \cdot \mu(\{x \mid |f(x)| \geq \lambda\}) \, d\lambda. $$Now, we split the integral into two parts: \begin{eqnarray*} ... 1 Your function produces infinities in both endpoints of the interval. The usual approach is to split the integral using some well-behaved intermediate point, say x=1. Therefore, \int \limits _0 ^\infty = \int \limits _0 ^1 + \int \limits _1 ^\infty. Concerning the first one, make the change of variable x = \Bbb e ^{-t}, obtaining \int \limits _0 ... 0 Using the inequality \frac{x}{x+1} < \log (1+x) < x for ll x>-1 and x \ne 0. Let x=u-1 then the inequality becomes \frac{u-1}{u} < \log (u) < u-1 for all u>0. Thus, \frac{(x-1)^2}{x^2}< (\ln x)^2<(x-1)^2 so that 1+\frac{(x-1)^2}{x^2}<1+ (\ln x)^2<1+(x-1)^2 which implies that ... 1 If \mu(X)=1, there exists p>0 with f\in L^p(\mu), and if we agree that \log(0)=-\infty and \exp(-\infty)=0 then$$\lim_{p\to0}||f||_p=\exp\left(\int\log|f|\,d\mu\right), $$the geometric mean of |f|. Hence some people sometimes write ||f||_0 for the geometric mean. Assume f\ge0. First, Jensen's inequality with \phi(t)=e^{pt} shows that ... 2 In general, the p-norms of a function aren't increasing. If there is a measurable set E with 1 < \mu(E) < +\infty, then$$\lVert \chi_E\rVert_p = \mu(E)^{\frac{1}{p}}$$is decreasing. Generally, there are f for which \lVert f\rVert_p is (eventually) decreasing and f for which \lVert f\rVert_p is (eventually) increasing. If \mu is a ... 0 Since you put no assumptions on u apart from L_{\rm loc}^1, the question could be recast as: what functions \omega are multipliers on L_{\rm loc}^1, in the sense that u\in L_{\rm loc}^1\implies u\omega \in L_{\rm loc}^1? The answer to which is: precisely the locally bounded functions, L_{\rm loc}^\infty. The argument is essentially the same as ... 3 Suppose f\in L^\infty([0,1]) and \|f\|_\infty> 0. Let E = \{x:|f(x)|=\|f\|_\infty\}. Then$$\lim_{p\to \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p = \frac{1}{m(E)}.$$(If m(E)=0, the conclusion is that the limit is \infty.) Proof: Let M= \|f\|_\infty. Then the expression equals$$\frac{M^p}{M^p\cdot m(E) + M^p\int_{[0,1]\setminus ...

4

You've done the bulk of the work already. To show that $\|f_{n_k} - f \|_{L^p(E \setminus F)}$ is bounded, note that by the triangle (Minkowski's) inequality, $$\|f_{n_k}-f \|_{L^p(E \setminus f)} \leq \|f_{n_k}\|_{L^p(E)} + \|f\|_{L^p(E)}.$$ By Fatou's lemma, $$\|f\|_{L^p(E)}^p = \int_E |f|^p = \int_E \lim\limits_{k \to \infty} |f_{n_k}|^p \leq ... 0 Whenever \Omega is a space of finite measure and f is in L^p(\Omega), it follows that f is in L^q(\Omega) for 1\leqslant q<p, for$$\lVert f\rVert_q \leqslant \mu(\Omega)^{1/q-1/p} \lVert f\rVert_p$$In particular for \Omega a probability space we get 1\leqslant q\mapsto \lVert f\rVert_q is an increasing function. Thus \lVert ... 2 For 0<p\le 1 by the c_r inequality$$|\mathbb{E}|X_n|^p-\mathbb{E}|X|^p|\le \mathbb{E}|X_n-X|^p\to 0$$For p>1 the Minkowski inequality implies that$$|||X_n||_p-||X||_p|\le ||X_n-X||_p\to 0$$3 X_n \to X in L^p for p>1 implies X_n \to X in L^1 because || X||_p is an increasing function in p (use Jensen), in particular EX_n \to EX. Since x \mapsto x^p is continuous this also implie s(EX_n)^p \to (EX)^p. 0 Think about a normed linear space. Any normed linear space. The norm is continuous since$$ \left|\|x_n\| - \|y_n\| \right| \le \|x_n - y_n\|.$$2 Let f(x) = \frac{1}{|x|} and g(x) = \chi_{|x|>1}\frac{1}{|x|}+\chi_{|x|\le 1}|x|... Note here that technically g(0)=0, but at least g>0 almost everywhere. 5 Indeed, usually one allows non-negative functions to take the value +\infty, but one usually doesn't talk about integrable functions taking values in \overline{\Bbb R}. There are good reasons for that; I shuddered when I saw that definition \infty-\infty=0. I hope his main point in doing this is to point out that it's not a good idea... Anyway. If the ... 1$$ |(f*g)(x)=\int f(x-y)g(y)\, dy| \leq \left( \int |f(x-y)|^p \, dy \right)^{1/p} \left( \int |g(y)|^q \, dy \right)^{1/q} = \|f\|_p \|g\|_q $$for (almost) all x. Hence \|f*g\|_\infty \leq... 6 This is simply false. \sin is locally square integrable, i.e. \sin \in L^2([a,b]) for all a<b\in\mathbb R. This can also be written as \sin \in L^2_{\text{loc}}(\mathbb R). To see that estimate$$\int_a^b \sin^2 x \ \mathrm dx \le \int_a^b 1 \ \mathrm dx = (b-a) < \infty$$Thus \sin \in L^2([a,b]) for any a<b\in\mathbb R, so we are done. ... 0 This, as you've stated, is not true: the function x \mapsto e^{-2\pi i x} is of course orthogonal to all the elements on the aforementioned set, and is nonzero. In order to obtain an orthonormal basis, you should've included \{ e^{-2\pi i k x}; k \in \Bbb{N} \} in the definition of the set above. To prove this set is really an orthonormal basis, you ... 0 As demonstrated here, the \sigma-finiteness assumption is not necessary for the "in measure" version of the dominated convergence theorem to hold. (Note that it happens to be the very link from your question yet Davide Giraudo edited his answer after PhoemueX's comment). 1 Yes, you've used it correctly. It's true that |\hat{f}| \le \|f\|_{L^\infty} only almost everywhere, but that is sufficient for your conclusion. To see why, let A = \{|\hat{f}| \le \|f\|_{L^\infty}\}, and break your integral into two integrals, over the sets A and A^c. Then note that A^c has measure zero. Intuitively, the Lebesgue integral ... 4 In fact \sin (n_kx) diverges a.e. for every subsequence n_k. Proof: Fix a subsequence n_k and let E= \{x\in \mathbb {R}: \lim_{k\to\infty}\sin (n_kx)\,\,\text {exists}\}. Let f be the pointwise limit function on E. Let a> 0 and put E_a=E\cap [-a,a]. On E_a we have f\in L^1\cap L^2. By the DCT,$$\int_{E_a}f^2 = \lim_{k\to\infty} ...

3

A different proof. If $\{\sin(n_k\,x)\}$ converges a.e., then $$(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2$$ converges to $0$ a.e. By the dominated convergence theorem $$\lim_{k\to\infty}\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=0,$$ but $$\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=1.$$

Top 50 recent answers are included