# The $L^1$ convergence of $f(x - a_n) \to f(x)$

I'm trying to solve something, and I'm stuck.

Suppose you have a function $$f \in L^1(\mathbf R)$$ and a sequence of real numbers that converges to zero: $$a_n \rightarrow 0$$ define a sequence of functions by $$f_n (x)=f(x-a_n)$$ does this sequence converges to $f(x)$ in the norm?

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So, you know that $f$ is integrable and hence you know that $|f_n(x) - f(x)| \leq 2|f(x)|$. So... –  Jonas Teuwen Sep 26 '11 at 12:48
So you say i can use now Lebesgue's bounded convergence theorem. Thanks :) –  Max Sep 26 '11 at 13:02
There are some things you must not forgot when applying this. When does $f(x - a_n) \to f(x)$? –  Jonas Teuwen Sep 26 '11 at 13:04
Also please note that it is $f \in L^1$, not $f(x) \in L^1$, $L^1$ is a function space and $f(x)$ is the value of the function $f$ at the point $x$. Something different. –  Jonas Teuwen Sep 26 '11 at 13:27
yes you are right. Thanks again for the answer! –  Max Sep 26 '11 at 13:54

So what you are trying to do is prove that

$$\int |f(x) - f_n(x)| \, \text{d}x \to 0.$$

Note that $|f(x) - f(x - a_n)| \to 0$ pointwise if $f$ is continuous with compact support. Furthermore, note that $f$ is uniformly continuous because it has compact support (and hence we can concatenate the zero function with a function on a bounded interval). So, given $\epsilon > 0$ we have $|f(x) - f_n(x)| < \epsilon$ for all $x$ and $n$ large enough. This means that for $n$ large enough we can find a compact support for the function $f - f_n$ so, we can bound this by an integrable function. So we can apply the Lebesgue Dominated Convergence Theorem to obtain the result for continuous $f$ with compact support.

Now we know that the compactly supported continuous functions are dense in $L^1(\mathbf R)$. So, let $f$ be a general $L^1$ function and let given $\epsilon > 0$, $g$ be a compactly supported continuous function such that $\|f - g\| < \epsilon$.

So, we get $\|f - f_n\|_1 \leq \|f - g\| + \|g - f_n\|$.

This requires us to show that $\|g - f_n\|$ can be made arbitrary small for large enough $n$. So given $\epsilon > 0$, let $g_n(x) = g(x - a_n)$. So,

$$\|g - f_n\| \leq \|g - g_n\| + \|g_n - f_n\|$$

The first term on the RHS goes to $0$ and the second one can be made arbitrarily small by density.

So, to solve these kinds of problems, first pick a class of functions that is dense in the required set where the problem becomes much easier.

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In general it is not true that $|f(x)-f_n(x)|\le2|f(x)|$, not even almost everywhere. However, if $f$ is continuous with compact support, it is uniformly continuous. Then, given any $\epsilon>0$, $|f(x)-f_n(x)|\le\epsilon$ for all $x$ and all $n$ large enough. Since there is a compact set containing the supports of $f$ and all $f_n$ for $n$ large enough, it follows that $\|f-f_n\|_1\to0$. The rest of the argument is correct. –  Julián Aguirre Sep 26 '11 at 14:22
@JuliánAguirre: Right, that is a stupid mistake, I will correct it. –  Jonas Teuwen Sep 26 '11 at 15:22