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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?

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1 Answer 1

up vote 3 down vote accepted

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 $M\cdot\lvert f\rvert$.

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Why is RHS uniformly dominated by $M|f|$? as $h$ gets really small.. wouldnt that be a problem? –  john Dec 2 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 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 at 21:53

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