Let $f\in L^p(0,1)$ and define $f_h$

Question

Let $f\in L^p(0,1)$ ($1\leq p<\infty$) and define $f_h$ as $$f_h(x)=\begin{cases}f(x+h)&\text{ for } x+h\in [0,1]\\ 0 &\text{ for } x+h\not\in[0,1]\end{cases}$$ Prove that for all $\varepsilon>0$ there is $\delta>0$ such that $|h|<\delta$ implies $\|f-f_h\|_p<\epsilon$.

This is a previous qual question. I was thinking of letting $\delta=\min\{\frac{\sqrt[p]{\varepsilon}}{\|f\|_p},1\}$ and breaking up the integral but I'm not sure how to proceed.

$$\|f_h-f\|_p^p=\int_0^1 |f_h-f|^p~dx=\int_0^h|f_h-f|^p~dx+\int_h^1|f_h-f|^p~dx$$

Does anyone have any ideas? I would like hints only, over full answers, please.

Answer 1

If $p<\infty$, I'd use the density of $C_c(0,1)$ in $L^p$, together with the fact that a continuous function with compact support is uniformly continuous.

Answer 2

First, I prove this where $f$ is continuous. If $f$ is continuous on $[0,1]$ then it is uniformly continuous on $[0,1]$ and is bounded. Let $\varepsilon>0$ be given. Then there is some $\delta_1>0$ such that for all $x_0\in [0,1]$, if $0<|x_0-x|<\delta_1$ then $|f(x_0)-f(x)|<\sqrt[p]{\varepsilon^p/2}$. Since $f$ is continuous on a closed interval, it is bounded by some $M$. Let $\delta=\min\{\delta_1,\frac{\varepsilon^p}{2M^p}\}$. Then if $|h|<\delta$:

\begin{align*}\|f_h-f\|_p^p&=\int_0^1|f_h-f|^p~dx\\ &=\int_0^{1-h}|f_h-f|^p~dx+\int_{1-h}^1|f_h-f|^p~dx\\ &=\int_0^{1-h}|f(x+h)-f(x)|^p~dx+\int_{1-h}^1|f(x)|^p~dx\\ &\leq\int_0^{1-h} \varepsilon^p/2~dx+\int_{1-h}^1 M^p~dx\\ &=\varepsilon^p(1-h)/2+hM^p\\ &\leq\varepsilon^p/2+\varepsilon^p/2\\ &=\varepsilon^p\end{align*}

So finally $\|f_h-f\|_p<\varepsilon$.

Now let $f$ be any function $f\in L^p$ and consider $\varepsilon>0$. Since continuous functions are dense in $L^p$ there is some $g$ such that $\|f-g\|_p<\varepsilon/3$. From the previous part, we know that $\|g-g_h\|_p<\varepsilon/3$. Now all we need to show is that $\|g_h-f_h\|_p<\varepsilon/3$. Note:

\begin{align*}\|g_h-f_h\|_p^p&=\int_0^1|g_h-f_h|^p~dx\\ &=\int_0^{1-h}|g(x+h)-f(x+h)|^p~dx\\ &=\int_h^1|g(x)-f(x)|^p~dx\\ &\leq\int_0^1 |g(x)-f(x)|^p~dx\\ &=\|g-f\|_p^p\end{align*}

Thus applying the triangle inequality we get $\|f-f_h\|_p<\varepsilon$.