Show that $\lim\limits_{n \to \infty} \int_{\Bbb R} f(x) g(x+n) dx=0$

Question

Problem: Let $f, g \in L^2(\Bbb R)$. Show that $$\lim\limits_{n \to \infty} \int_{\Bbb R} f(x) g(x+n) dx=0$$

Solution: Consider $\Bbb R = S_n \cup T_n$ where: $$S_n =\{ x \ | \ |x| \leq \frac{n}{2} \} \text{ and } T_n =\{ x \ | \ |x| > \frac{n}{2}\}$$ $$ \int_{\Bbb R} f(x) g(x+n)dx = \int_{S_n} f(x) g(x+n) dx + \int_{T_n} f(x) g(x+n) dx$$

By Holder inequality and translation invariant property \begin{align*} \Bigg|\int_{S_n} f(x) g(x+n) dx \Bigg| & \leq \|f\|_2 \Big( \int_{S_n} g(x+n)^2 dx \Big)^{\frac{1}{2}}\\ & \leq \|f\|_2 \Big( \int_{|x|>\frac{n}{2}} (g(x))^2 dx \Big)^{\frac{1}{2}}\\ & =\|f\|_2 \cdot \epsilon\\ &=\epsilon \end{align*} because $\Big| \int_{|x| > \frac{n}{2}} g(x)^2 dx \Big| < \epsilon$

Same argument for $T_n$, therefore the whole thing is less than $\epsilon$.

Is this solution correct? Is there any other way of solving this problem Thanks

Answer 1

Your argument seems correct. Maybe it could be better to first declare $\varepsilon$, then choose $n_0$ such that $n\geqslant n_0$, $\int_{|x| > \frac{n}{2}} g(x)^2 dx\lt \varepsilon$ and the same for $f$.

• Unless you did not specify that the norm of $f$ is one, you can only bound by $\|f\|_2 \cdot \epsilon$.
• Consequently, the final bound is $\varepsilon\lVert f\rVert_2+\varepsilon\lVert g\rVert_2$.