4
$\begingroup$

I have a function $f\in L^1(\mathbb{R}) $ and $g(x)=\dfrac{1}{2\sqrt{\pi t}}e^{-\frac{(at+x)^2}{4t}}$, where $a,t\in\mathbb{R}$, $t>0$. I want to show that $$\dfrac{d}{dx}\int_{-\infty}^\infty f(y)g(x-y)dy=\int_{-\infty}^\infty f(y)\dfrac{d}{dx}g(x-y)$$ Leibniz doesn't work since there's no continuity assumption on $f$. I'm thinking about using the dominated convergence theorem, but how would the proof go?

$\endgroup$
4
$\begingroup$

Since you multiply it with an $L^1$ function, it is sufficient to show that $g'(x)$ is bounded, say $\lvert g'(x)\rvert \leqslant M$.

Then the dominated convergence theorem can be applied to the difference quotients

$$\frac{(f\ast g)(x+h) - (f\ast g)(x)}{h} = \int_{-\infty}^\infty f(y) \frac{g(x+h-y)-g(x-y)}{h}\,dy,$$

which then are uniformly dominated by $M\cdot\lvert f\rvert$.

$\endgroup$
  • $\begingroup$ Why is RHS uniformly dominated by $M|f|$? as $h$ gets really small.. wouldnt that be a problem? $\endgroup$ – john Dec 2 '14 at 21:45
  • 1
    $\begingroup$ Because $$\frac{g(x+h-y) - g(x-y)}{h} = g'(x+\theta\cdot h - y)$$ for some $\theta\in (0,1)$ by the mean value theorem, and since $g'$ is uniformly bounded by $M$, the integrand is dominated by $M\cdot\lvert f\rvert$. $\endgroup$ – Daniel Fischer Dec 2 '14 at 21:48
  • $\begingroup$ Thanks! Can you look at this post?math.stackexchange.com/questions/1048865/… Can I use the same MVT argument in this multidimensional case? $\endgroup$ – john Dec 2 '14 at 21:53
  • $\begingroup$ If $f$ is instead bounded (not necessarily integrable), how would you find the dominating function? $\endgroup$ – yumiko Feb 3 '17 at 20:32
  • 1
    $\begingroup$ @yumiko Although $\theta$ depends on $x,y$ and $h$, we always have $0 < \theta < 1$. So $x + \theta h$ lies between $x$ and $x+h$. Our dominating function works if $\lvert x + \theta h\rvert < K$, and that is guaranteed if $\lvert x\rvert < K$ and $\lvert x+h\rvert < K$. Since we're interested in $h \to 0$, we can restrict our attention to $\lvert h\rvert \leqslant 1$, and to see that we can differentiate under the integral at $x_0$, it suffices to choose $K > \lvert x_0\rvert + 1$. $\endgroup$ – Daniel Fischer Feb 3 '17 at 21:03

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.