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For $a\in R^d$, let $T_af(x)=f(x-a),$ for all $f\in L^p(R^d), 1\leq p<\infty$ and all $x\in R^d$. I need example that this operator doesn't converge uniformly when $a\rightarrow 0$. I know that this operator has strong (norm) limit.

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Let $d=1,p=1$ and $f_n(x)=nx^{n}$ for $0<x<1$, $0$ for $x \notin (0,1)$. Then $\|f_n\| <1$ for all $n$ . Suppose $\int_{\mathbb R}|f_n(x-a)-f_n(x)|\, dx \to 0$ uniformly in $n$ as $a\to 0$. Then, given $\epsilon \in (0,e^{-1}-e^{-2}) $ there exists $b>0$ such that $\int_{\mathbb R}|f_n(x-a)-f_n(x)|\, dx <\epsilon $ for all $a \in (0,b)$ for all $n$. Put $a=\frac 1 n$ where $n$ is so large that $\frac 1 n <b$. After a little computation we get $\frac n {n+1} ((1-\frac 1n)^{1+n}-(\frac 1 n)^{1+n}-(1-\frac 2 n)^{1+n}) <\epsilon $. Letting $n \to \infty$ we get $e^{-1}-e^{-2}<\epsilon$. This is a contradiction.

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  • $\begingroup$ Why you said that $sup$ converge to $\infty$ for each $a\in(0,1)$ when $a$ need to converge to 0? $\endgroup$ – Hana Jan 15 at 7:25
  • $\begingroup$ If we had uniform convergence then there would be $a_0$ such that $\int |f_n(x-a)-f_n(x)|\, dx <1$ for all $n$ for all $a<a_0$. In my example you can get a contradiction to this by fixing $a$, say $a=a_0/2$ and letting $n \to \infty$. $\endgroup$ – Kavi Rama Murthy Jan 15 at 7:28
  • $\begingroup$ Thanks, I understand that now. But I cannot get that limit is infinity: $\int_a^{1-a}|n(x-a)^n-nx^n|= \frac{n}{n+1}((1 - a)^{1+n} - a^{1 + n}-(1-2a)^{1+n})$ this converge to 0 for $a\in(0,1)$ when $n\rightarrow\infty$ or I made some mistake. $\endgroup$ – Hana Jan 15 at 16:03
  • $\begingroup$ @Hana I was a bit careless with details but the sequence $\{f_n\}$ does work. Please look at my revised answer. $\endgroup$ – Kavi Rama Murthy Jan 15 at 23:21
  • $\begingroup$ Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$? $\endgroup$ – Hana 2 days ago

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