Moving the derivative under the integral of a convolution I have functions $f(x)$ and $g(t,x)$, where $x \in \mathbb{R}^d; t \in \mathbb{R}$. I further assume that these functions fall off fast enough for $|x| \rightarrow \infty$. I want to show the following equality:
$$\int_{\mathbb{R}^d} (f(x) * \partial_t g(t,x) ) g(t,x) dx = \int_{\mathbb{R}^d} (f(x) * g(t,x) ) \partial_tg(t,x) dx.$$
However, I'm not sure what properties of the convolution product I have to use to show this.
 A: $$\int_{\mathbb{R}^d} \big(f(\cdot)*\partial_t g(t,\cdot)\big)(x)g(t,x)dx = \big\langle (f(\cdot)*\partial_t g(t,\cdot))(x),g(t,x)\big\rangle$$ $$= \left\langle \widehat{f(\cdot)*\partial_t g(t,\cdot)}(\xi),\widehat{g(t,\cdot)}(\xi)\right\rangle = \left\langle \widehat{f(\cdot)}(\xi)\widehat{\partial_t g(t,\cdot)}(\xi),\widehat{g(t,\cdot)}(\xi)\right\rangle$$ $$= \left\langle \widehat{f(\cdot)}(\xi)\widehat{g(t,\cdot)}(\xi),\widehat{\partial_t g(t,\cdot)}(\xi)\right\rangle = \int_{\mathbb{R}^d} \big(f(\cdot)*g(t,\cdot)\big)(x)\partial_t g(t,x)dt$$
A: It is false in general. If $f$ or $g$ is an even function of $x$, then it is true. You can just use the definition of the convolution and Fubini theorem to get
$$
\begin{align*}
\int_{\mathbb{R}^d} (f*\partial_tg(t,\cdot))(x)\, g(t,x)\,\mathrm{d} x &= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} f(x-y)\,\partial_tg(t,y)\,\mathrm{d}y\right) g(t,x)\,\mathrm{d} x
\\
&= \iint_{\mathbb{R}^d} f(x-y)\,\partial_tg(t,y)\, g(t,x)\,\mathrm{d} x \,\mathrm{d}y
\\
&= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} f(x-y)\,g(t,x)\,\mathrm{d}x\right) \partial_tg(t,y)\,\mathrm{d} y
\end{align*}
$$


*

*And now if $f$ is even then $f(x-y) = f(y-x)$, and you get
$$
\begin{align*}
\int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} f(y-x)\,g(t,x)\,\mathrm{d}x\right) \partial_tg(t,y)\,\mathrm{d} y &= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} f(x-y)\,g(t,x)\,\mathrm{d}x\right) \partial_tg(t,y)\,\mathrm{d} y
\\
&= \int_{\mathbb{R}^d} (f*g(t,\cdot))(x)\, \partial_tg(t,x)\,\mathrm{d} x
\end{align*}
$$

*If $g$ is even then you can do the changes of variables replacing $x$ by $-x$ and $y$ by $-y$, which will not change $g$ and $\partial_t g$ but will replace $f(x-y)$ by $f(y-x)$ so you get the same result.


And for it to be valid, you should either have positive functions, or check that the double integral $\iint_{\mathbb{R}^d} |f(x-y)\,\partial_tg(t,y)\, g(t,x)|\,\mathrm{d} x \,\mathrm{d}y$ is finite, or use distribution theory.

To understand why it is false in the general case, we can for example take $f = \delta_1$ in dimension $d=1$  to simplify (but you could also take a $C^\infty_c(\mathbb{R}^d)$ function with support in a very small ball centered around some unit vector). Then your equality would lead to
$$
\begin{align*}
∫_{\mathbb{R}^d} \partial_t g(x-1)\, g(x)\,\mathrm{d} x &= \int_{\mathbb{R}^d} g(x-1)\, \partial_tg(x)\,\mathrm{d} x
\\
&= \int_{\mathbb{R}^d} \partial_tg(x+1)\,g(x)\,\mathrm{d} x
\end{align*}
$$
which is false in general (take for example at some time $t$, $g$ positive compactly supported in $[-1/2,1/2]$ and $\partial_t g$ positive compactly supported in $[1/2,3/2]$, then the first integral is positive but the second is $0$.

You can also prove it with the Fourier transform, as done by Mathworker, but his proof is false since there should be a minus sign appearing somewhere. For example
$$
\begin{align*}
∫_{\mathbb{R}^d} (f*\partial_t g)(x)\, g(x)\,\mathrm{d} x &= \int_{\mathbb{R}^d} \widehat{f*\partial_t g}(x)\, \widehat{g}(-x)\,\mathrm{d} x
\\
&= \int_{\mathbb{R}^d} \widehat{f}(x)\,\partial_t \widehat{g}(x)\, \widehat{g}(-x)\,\mathrm{d} x
\\
&= \int_{\mathbb{R}^d} \widehat{f}(-x)\,\partial_t \widehat{g}(-x)\, \widehat{g}(x)\,\mathrm{d} x
\\
&= \int_{\mathbb{R}^d} \widehat{\check{f}*g}(x)\, \widehat{\partial_tg}(-x)\,\mathrm{d} x
\\
&= \int_{\mathbb{R}^d} \check{f}*g(x)\, \partial_tg(x)\,\mathrm{d} x
\end{align*}
$$
where $\check{f}(x) = f(-x)$.
