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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?

Kunal
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1 Answers1

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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 $M\cdot\lvert f\rvert$.

Daniel Fischer
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  • Why is RHS uniformly dominated by $M|f|$? as $h$ gets really small.. wouldnt that be a problem? – john Dec 02 '14 at 21:45
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    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 02 '14 at 21:48
  • Thanks! Can you look at this post?http://math.stackexchange.com/questions/1048865/convolution-with-c-infty-produces-c-infty Can I use the same MVT argument in this multidimensional case? – john Dec 02 '14 at 21:53
  • If $f$ is instead bounded (not necessarily integrable), how would you find the dominating function? – yumiko Feb 03 '17 at 20:32
  • @yumiko With the $g$ from the question, one uses $g'$. Disregarding constant factors, $g'(x) = x\exp\bigl(-c(d+x)^2\bigr)$, and so for $\lvert x\rvert < K$ we have $\lvert g'(x-y)\rvert \leqslant C\bigl\lvert (K+\lvert y\rvert) \exp\bigl(-cy^2 + 2c(K+\lvert d\rvert)\lvert y\rvert\bigr)\bigr\rvert$, which is integrable (with respect to $y$). That decays fast enough that we don't even need $f$ bounded, $f$ just mustn't grow too fast. For other $g$, there may or may not be a dominating function for $f(y)g'(x-y)$. – Daniel Fischer Feb 03 '17 at 20:45
  • But, how do you deal with $\theta$? Don't I need to bound $g'(x+\theta h -y )f(y)$ where $\theta$ depends on $y$? – yumiko Feb 03 '17 at 20:54
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    @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 03 '17 at 21:03
  • I am still trying to digest what you said, and am I correct to say that you are essentially bounding $g'$ by itself? – yumiko Feb 03 '17 at 21:49
  • @yumiko Not quite by itself, but by a modification of itself. Essentially, let $$h_M(x) = \sup ; { \lvert g'(t)\rvert : \lvert t-x\rvert \leqslant M},$$ then the exponential decay of $g'$ ensures that $h_M$ is integrable for all $m\in (0,+\infty)$, and it dominates $g'(x + \theta h)$ for $\lvert h\rvert < M$. – Daniel Fischer Feb 03 '17 at 21:56