$L_{p}$ 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?

Solution 1:

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 $|t| \lt \delta(g,\varepsilon)$.