$L_{p}$ distance between a function and its translation

I'm working through a proof and one of the comments is that for a function $f\in L_p (\mathbb{T})$:

$$\lim_{t\to 0}\;\|f(\cdot + t) - f\|_p = 0.$$

How do I prove it? I think it is intuitively clear if $f$ is a step function, but what about for an arbitrary $p$ integrable function?

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Prove it for continuous functions first, using that they are uniformly continuous, then use that they are dense. See Jonas's answer here for the $L^1$-case and adapt the argument to $L^p$, $p \lt \infty$. And yes, it should be $t \to 0$. – t.b. Oct 4 '11 at 3:15

1 Answer

Note that the restriction $1 \leq p \lt \infty$ is necessary here. For $p = \infty$ just consider a characteristic function of a proper subinterval of $\mathbb{T}$. Your idea with characteristic functions can be made into an argument for $p \lt \infty$ but the following seems simpler to me:

1. Since $\mathbb{T}$ is compact, every continuous function is uniformly continuous. This means: for every continuous function $g$ and every $\varepsilon \gt 0$ there is $\delta = \delta(g,\varepsilon) \gt 0$ such that for all $|t| \lt \delta$ the estimate $|g(x+t) - g(x)| \lt \varepsilon$ holds. Integrating this over $\mathbb{T}$ we see that $\|g(\cdot+t) - g\|_p \leq \varepsilon$ for all $|t|\lt \delta$.

2. Now for every $f \in L^p(\mathbb{T})$ and $\varepsilon \gt 0$ there is a continuous $g$ such that $\|f-g\|_p \lt \varepsilon$. Using 1., this gives $$\|f(\cdot+t)-f\|_p \leq \|f(\cdot+t) - g(\cdot+t)\|_p + \|g(\cdot+t)-g\|_p + \|g-f\|_p \leq 3\varepsilon$$ for all $|t| \lt \delta(g,\varepsilon)$.

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Thank you very much. I feel pretty thick for not being able to answer these questions myself. Your explanation makes it seem dead obvious. Thank you. – roo Oct 4 '11 at 3:55
Don't worry too much, Kyle, one needs to get the hang of this, and it takes some time... I saw your question in which you said that you're lagging behind the course and when you're feeling under pressure the mind usually can't focus as well as it usually does. Keep your head up and use the site for what it's for, good luck! – t.b. Oct 4 '11 at 4:07