Tag Info

Hot answers tagged

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. ...


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 ...


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 ...


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 ...


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 ...


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 ...


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$(EX_n)^p \to (EX)^p$.


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 $$|||X_n||_p-||X||_p|\le ||X_n-X||_p\to 0$$


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

Let $f_n$ be a square function with width $1/n$ and height $\sqrt{n}$. Then $f_n\rightarrow 0$ in $L^2$. Let $g_n=f_n$...


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.


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 ...


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 ...


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} ...


2

You are almost right. Just use the Cauchy-Schwarz inequality $$\mu(\Omega)=\int_{\Omega} 1 \,d\mu=\int_{\Omega} \left|f^{1/2}\frac{1}{f^{1/2}}\right|\,d\mu\leqslant \left(\int_{\Omega}|f|\, d\mu\right)^{\frac{1}{2}}\left(\int_{\Omega}\frac{1}{|f|}\, d\mu\right)^{\frac{1}{2}} $$


2

Both results follow immediately with an application of Young's inequality Let $1 \le p,q,r \le \infty$ be such that $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1.$$ If $f \in L^p$ and $g \in L^q$ then $f * g \in L^r$ and we get the following bound on the norm $$\|f * g\|_r \le \|f\|_p\|g\|_q.$$ To answer your first question, let $f \in L^1$ and ...


1

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

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).


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 ...


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 $L^q \subset L^p$.


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 ...


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 $L^p\times L^q\to\mathbb F$.


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 $\|f*g\|_\infty \leq$...


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 ...


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*} ...



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