If $f$ is a continuous function with compact support, then $\lim\limits_{h\to\infty}\int_{\mathbb{R}}|f(x-h)+f(x)|dx=2\int_{\mathbb{R}}|f(x)|dx$

Question

Here is the problem:

If $f$ is a continuous function with compact support on $\mathbb{R}$, prove that $\displaystyle\lim\limits_{h\to\infty}\int_{\mathbb{R}}|f(x-h)+f(x)|dx=2\int_{\mathbb{R}}|f(x)|dx$.

My idea is to separate $\mathbb{R}$ into two parts: $|x-h|<M$ and $|x-h|>M$, where $M>0$ is chosen such that $\forall x\in supp(f)$ satisfies $|x|<M$, and then try to deal with the two parts of integration, but I found it hard to deal with this part: $\int_{|x-h|<M}|f(x-h)+f(x)|dx$.

Does there exsit a better method? Or how to complete the proof in the way above?

When $f$ is only continuous on $\mathbb{R}$, does the conclusion still hold? (I guess it doesn't hold)

Answer 1

Suppose $f(x)=0$ for $|x| >M$. If $h >2M$ then $f(x)=0$ or $f(x-h)=0$ for every $x$. Indeed either $|x| >M$ or $|x-h| >M$ because on the contrary, $h=|(x-h)-x| \leq |x-h|+|x|<2M$. Hence $|f(x-h)|+|f(x)|=|f(x-h)+f(x)|$ Now integrate.

There is no better method.

Actually the result is true for any integrable function $f$. This follows easily from above result and the fact for every $\epsilon >0$ there exist a continuous function $g$ with compact support such that $\int_{\mathbb R} |f(x)-g(x)|dx <\epsilon$.