1
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Ant Dec 18 '14 at 10:57
  • $\begingroup$ That wasn't covered in the lectures so probably not $\endgroup$ – D. Vente Dec 18 '14 at 11:00
2
$\begingroup$

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 $L^1$.)

$\endgroup$
1
$\begingroup$

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 )

$\endgroup$
  • $\begingroup$ 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 $f(s)$ $\endgroup$ – D. Vente Dec 18 '14 at 14:54
  • $\begingroup$ 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 $\endgroup$ – Ant Dec 18 '14 at 15:07
  • 1
    $\begingroup$ 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? $\endgroup$ – D. Vente Dec 18 '14 at 15:08
  • $\begingroup$ @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! :-) ) $\endgroup$ – Ant Dec 18 '14 at 15:11
  • 1
    $\begingroup$ 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 $\mathbb{R}$) $\endgroup$ – mrf Dec 18 '14 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.