# Prove that a intergral over $\mathbb R$ is finite

Let $K\in \mathcal L_1(\mathbb R)$ and $f$ be measurable and bounded on $\mathbb R$ such that $\lim_{|x|\to \infty} f(x)=0$. Define $$F(x):= \int _{\mathbb R} K(x-s)f(s)\;ds \qquad (x\in \mathbb R)$$ Prove that $F(x)$ is finite for all $x\in \mathbb R$ and that $\lim_{|x|\to \infty} F(x)=0$. I was told that the proof would probably include the use of the Fubini's theorem or Tonelli's Theorem, but I don't see how they apply here...

• Note that $F = K * f$, so if you know the properties of convolution it should be easy. You are probably not allowed to use them though – Ant Dec 18 '14 at 10:57
• That wasn't covered in the lectures so probably not – D. Vente Dec 18 '14 at 11:00

For the first part, since $f$ is bounded: $$|F(x)| = \left|\int_{\mathbb R} K(x-s)f(s)\,ds\right| = \left| \int_{\mathbb R} f(x-s)K(s)\,ds \right| \le M \int_\mathbb R |K(s)|\,ds$$ where $M = \sup_\mathbb{R} |f|$. (The absolute values are missing in @Ant's solution.)

For the second part, \begin{align} \lim_{x \to \pm\infty} F(x) &= \lim_{x \to \pm\infty} \int_{\mathbb{R}} K(x-s)f(s)\,ds \\ &= \lim_{x \to \pm\infty} \int_{\mathbb{R}} f(x-s)K(s)\,ds \\ &= \int_{\mathbb{R}} \lim_{x \to \pm\infty} f(x-s)K(s)\,ds = 0 \end{align} by dominated convergence. (Note that $|f(x-s)K(s)| \le M |K(s)|$ for all $x$, and the right hand side in the inequality is $L^1$.)

Note that $f$ is bounded on $\mathbb R$, so set $M = \sup |f(x)|$, then $$F(x) = \int_\mathbb R K(x-s)f(s)ds = \int_\mathbb Rf(x-s)K(s)ds\le M \int_\mathbb R K(s)ds$$ and since $K \in \mathcal L^1(\mathbb R)$ you are done.

To show that $\lim_{x \to \infty} F(x) = 0$, note that for every $\varepsilon > 0$, exists $N > 0$ such that for every $x > N$, $|f(x)| < \varepsilon$. But this also implies that $$\left|\int_\mathbb R K(x-s)f(s)ds\right| \le \int_\mathbb R \left | K(x-s)f(s)ds\right| = \int_\mathbb R |K(x-s)| \cdot |f(s)| ds \le \varepsilon \int_\mathbb R |K(x-s)|ds$$

Since $K \in \mathcal L^1(\mathbb R)$, set $L = \int_\mathbb R |K(x-s)|ds \in \mathbb R$. This is the exact definition of $\lim_{x \to \infty} F(x) = 0$ !

That is to say, for every $\varepsilon > 0$ exists $N$ such that for every $x > N$ we have $$|F(x)| = \left|\int_\mathbb R K(x-s)f(s)ds\right| \le \varepsilon L$$

(note that the fact that $\varepsilon$ is multiplied by a constant does not change anything )

• Here you use the fact that $f(s)<\varepsilon$ however you have a $x$ such that $f(x)<\varepsilon$. We don't know anything about $f(s)$ – D. Vente Dec 18 '14 at 14:54
• Do not be confused with the letters! $f(x)$ is a function, its variable is $x$. But $f(s)$ is the same function, this time the variable is $s$! If I tell you that $f(x) = e^{-x^2}$ is going to $0$ as $x \to \infty$, do you think that the behaviour of $f(s) = e^{-s^2}$ will be different? Since both $x$ and $s$ are independent variables writing $f(x)$ or $f(s)$ is the exact same thing – Ant Dec 18 '14 at 15:07
• They are but sine you integrate over whole $\mathbb R$ you must also consider the behaviour of $f$ where $x\leq N$, doesn't the argument fall apart there? – D. Vente Dec 18 '14 at 15:08
• @D.Vente You just need to consider $x \to -\infty$, not on the whole real line. And for $x \to -\infty$ one can make the exact same argument, because $\lim_{x \to \pm \infty} f(x) = 0$ (changing some signs of course, but it is pretty much the same thing. You may want to try do write it down yourself! :-) ) – Ant Dec 18 '14 at 15:11
• The last inequality in the second displayed formula looks suspicious. ($|f(s)|$ is only small when $|s|$ is large, and you are integrating over all of $\mathbb{R}$) – mrf Dec 18 '14 at 15:19