If $f$ is $L^1$ and continuous, then does $\int |f(a-x)-f(a)|dx<\infty$ hold?

Question

Let $f:\mathbb R\to \mathbb C$ be a $L^1$ and continuous function, and $a\in\mathbb R.$

Then, does $$\int_{-\infty}^\infty |f(a-x)-f(a)|dx<\infty$$ hold ?

Now, I have $\int_{-\infty}^\infty |f(x)|dx<\infty.$

And the continuity of $f$ at $a$, for all $\epsilon,$ I get $\delta>0$ s.t. $|y-a|<\delta \implies |f(y)-f(a)|<\epsilon.$

So \begin{align} &\int_{-\infty}^\infty|f(a-x)-f(a)|dx\\ &=\int_{|x|<\delta}|f(a-x)-f(a)|dx+\int_{|x|\geqq\delta}|f(a-x)-f(a)|dx\\ &<\int_{|x|<\delta}\epsilon dx+\int_{|x|\geqq\delta}|f(a-x)-f(a)|dx. \end{align}

$\int_{|x|<\delta}\epsilon dx$ is finite.

Can I find the bound for $\int_{|x|\geqq\delta}|f(a-x)-f(a)|dx$ ?

Answer 1

No, that integral is never finite except if $f(a) = 0$.

If $f(a) \ne 0$ then $$ |f(a-x)-f(a)| \ge |f(a)| - |f(a-x)| $$ implies that $\int_{-\infty}^\infty |f(a-x)-f(a)| \, dx = \infty$.

Answer 2

it is not true. Take $f(x) =e^{-x^2}$that is in $L^1$ for Lebesgue measure. Take $a=0$. Then

$$f(a-x) =f(-x) =f(x) $$ With parity property.

Hower $f(a) =1$. You see that $f(a-x)-f(a) $ tends to 1 as $x$ tends to infinity. It is no more $L^1$