# Is $(df(t)/dx) g(t)\star f(t)$ equal to $f(t) g(t)\star(df(t)/dx)$?

i tried to prove： $$\dfrac{df(t)}{dx} \cdot g(t)\star f(t) = f(t)\cdot g(t)\star \dfrac{df(t)}{dx}$$ where $$\star$$ is convolution, $$f$$ is both a function of $$x$$ and a function of $$t$$, and $$g$$ is just a function of $$t$$.

I think this equation holds, I tried the following: \begin{align} L.H.S =& \dfrac{df(t)}{dx} \cdot \int_{-\infty}^{\infty} g(\tau)f(t-\tau)d\tau \\ =& \int_{-\infty}^{\infty} g(\tau)f(t-\tau) \dfrac{df(t)}{dx} d\tau \end{align}tau and \begin{align} R.H.S =& f(t) \cdot \int_{-\infty}^{\infty} g(\tau)\dfrac{df(t-\tau)}{dx}d\tau \\ =& \int_{-\infty}^{\infty} g(\tau)f(t) \dfrac{df(t-\tau)}{dx} d\tau \end{align} If you use direct variable substitution, that is $$t=t-\tau$$, then you can prove that the equation is true, but $$t$$ is not an integral variable. I think this variable substitution cannot be used, but I can't find any other way to prove it.

How should I prove that the equation is true?

## 1 Answer

The identity you want to prove is not true in general. As a counterexample, let's consider $$f(x,t)=xt+x^2t^2$$ and $$g(t)=e^{-|t|}$$: \begin{align} \partial_xf(x,t)\,g(t)\star f(x,t)&=(t+2xt^2)\int_{-\infty}^{\infty}e^{-|\tau|}\left[x(t-\tau)+x^2(t-\tau)^2\right]d\tau \\ &=(t+2xt^2)\left[2xt+2x^2(t^2+2)\right], \tag{1} \\ f(x,t)\,g(t)\star\partial_xf(x,t)&=(xt+x^2t^2)\int_{-\infty}^{\infty}e^{-|\tau|}\left[(t-\tau)+2x(t-\tau)^2\right]d\tau \\ &=(xt+x^2t^2)\left[2t+4x(t^2+2)\right], \tag{2} \end{align} hence $$\partial_xf(x,t)\,g(t)\star f(x,t)-f(x,t)\,g(t)\star\partial_xf(x,t)=-4x^2t. \tag{3}$$