# Differentiating under integral for convolution

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?

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 $M\cdot\lvert f\rvert$.

• Why is RHS uniformly dominated by $M|f|$? as $h$ gets really small.. wouldnt that be a problem? – john Dec 2 '14 at 21:45
• 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 $M\cdot\lvert f\rvert$. – Daniel Fischer Dec 2 '14 at 21:48
• Thanks! Can you look at this post?math.stackexchange.com/questions/1048865/… Can I use the same MVT argument in this multidimensional case? – john Dec 2 '14 at 21:53
• If $f$ is instead bounded (not necessarily integrable), how would you find the dominating function? – yumiko Feb 3 '17 at 20:32
• @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 $K > \lvert x_0\rvert + 1$. – Daniel Fischer Feb 3 '17 at 21:03