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I came across a text that proves that translation operator $T_a(f):=f(x-a)$ where $a\in\mathbb{R}^n$ and $f\in L^p(\mathbb{R}^n)$ is continuous. The proof follows: $$||f(x-a)-f(x)||_p=||f(x-a)-g(x-a)+g(x-a)-g(x)+g(x)-f(x)||_p\leq ||f(x-a)-g(x-a)||_p+||g(x-a)+g(x)||_p+||g(x)-f(x)||_p<3\varepsilon$$ Where $g$ is some continuous function and hence $C^\infty$ is dense in $L^p$ the inequality holds. Wouldn't that meant $f$ is also continuous which it doesn't have to be? Why does the proof hold, respectively what is the catch with the proof, why isn't it the proof of continuity of $f$?

user74200
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  • Question for the poster: It looks to me like that inequality is trying specifically to show $T_a(f) \rightarrow f$ as $a \rightarrow 0$. Is that true? Are you showing continuity in the parameter or continuity for fixed $a$ as an operator? – Jason Knapp Nov 12 '14 at 16:40
  • Well, I am not showing anything, the author of the text is showing something and I don't know what is he trying to show. It is different from a definition of continuity I know and it does not make strong sense to me. – user74200 Nov 12 '14 at 16:46
  • I'm sorry to prod you with questions, but are you inferring "$f$ is continuous" from "$f_n \rightarrow f$"? I'm still not seeing where you are drawing the conclusion that $f$ is continuous. – Jason Knapp Nov 12 '14 at 16:46
  • Sorry, no, I mixed things up. – user74200 Nov 12 '14 at 16:50

2 Answers2

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First of all, in the proof you need to assume something else about the function $g$. Usually one takes $g\in C_c(\mathbb{R}^n)$, the space of continuous functions with compact support. This implies that $g$ is uniformly continuous, allowing one to prove that $\|g(x-a)-g(x)\|_p\to0$ as $a\to0$.

The proof shows that $$ \lim_{a\to0}\int_{\mathbb{R}^n}|f(x-a)-f(x)|^p\,dx=0. $$ This is very different of continuity of $f$ , which would be $$ \lim_{a\to0}|f(x-a)-f(x)|=0\quad\forall x\in\mathbb{R}^n. $$

Julián Aguirre
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Your statement in the comments that an operator $T$ is continuous on a Banach space $X$ if, for a sequence $f_n \rightarrow f$ in $X$ then we have $Tf_n \rightarrow Tf$. That is about the operator $T$ being continuous. The functions $f_n$ and $f$ are simply an arbitrary collection of elements in the space that form a convergent sequence and its limit.

Consider that the definition of a continuous operator makes sense even on a space where the elements are not functions themselves!

Jason Knapp
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