# Uniform convergence for operator of translation

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.

Let $$d=1,p=1$$ and $$f_n(x)=nx^{n}$$ for $$0, $$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 . 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.
• Why you said that $sup$ converge to $\infty$ for each $a\in(0,1)$ when $a$ need to converge to 0? – Hana Jan 15 at 7:25
• 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 $n \to \infty$. – Kavi Rama Murthy Jan 15 at 7:28
• 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. – Hana Jan 15 at 16:03
• @Hana I was a bit careless with details but the sequence $\{f_n\}$ does work. Please look at my revised answer. – Kavi Rama Murthy Jan 15 at 23:21
• Thanks! Whether the uniform convergence of this operator $T_a$ depends on p? I.e. whether the uniform convergence will be valid for some $p$? – Hana Jan 17 at 19:25