Proof of derivative of convolution $(h * f)' = h' * f$

Question

I am writing an article related to Computer Vision but to complete the math under the technique I should prove the following equality:

$$\frac{\mathrm d}{\mathrm dx}(h*f) = \left(\frac{\mathrm d}{\mathrm d x}h \right)*f$$

I am not a mathematician and I am not used to convolution analysis. So, how can I prove it?

You can start with the definition of convolution. – lisyarus Oct 31 '19 at 15:55 — lisyarus, Oct 31 '19 at 15:55
X-posted: https://dsp.stackexchange.com/q/61627/21075 – Rodrigo de Azevedo Oct 31 '19 at 17:10 — Rodrigo de Azevedo, Oct 31 '19 at 17:10

score 5 · Accepted Answer · answered Oct 31 '19 at 16:01

Recall that the convolution is defined as $$(h\ast f)(x) = \int_{-∞}^∞ h(x - τ)f(τ) \mathrm d τ \quad \text{for $x ∈ ℝ$}.$$ Now use the special case of Leibniz’s integral rule.

To apply this, you only need $h$ to be continuously differentiable.

score 1 · Answer 2 · edited Apr 21 '20 at 22:44

1

\begin{align} \frac{d}{d x}(h*f) &= \frac{d}{d x}\int f(t)h(x-t)dt\\ &= \int \frac{\partial}{\partial x}\left(f(t)h(x-t)\right)dt \\ &= \int f(t)\frac{\partial h(x-t)}{\partial x}dt \\ &= \left(\frac{d}{d x}h\right)*f \end{align}

edited Apr 21 '20 at 22:44

Manolito Pérez

1,241

answered Oct 31 '19 at 16:00

Ishan Deo

3,574

The step from the first to the second line is very much nontrivial – univalence Dec 08 '23 at 17:02

Proof of derivative of convolution $(h * f)' = h' * f$

2 Answers2