discrete signals - For an LTI system, why does the Fourier transform of the impulse response give the frequency response?

I know that for a given system, the Fourier transform of its impulse response gives its frequency response. I want to find where this property comes from, but haven't been able to find if it's a definition or if there's a mathematical proof available, for both continuous and discrete-time systems.

Answer

Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$, the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$ In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is $$\begin{align} y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\ &= \exp(j2\pi ft)\int_{-\infty}^\infty h(\tau)\exp(-j2\pi f\tau)\,\mathrm d\tau\\ &= H(f)\exp(j2\pi ft),\tag{1} \end{align}$$ where $H(f)$ is the Fourier transform of the impulse response $h(t)$. In words, for an LTI system with impulse response $h(t)$, the input $\exp(j2\pi ft)$ produces output $H(f)\exp(j2\pi ft)$. This is precisely what we have as the definition of the frequency response of an LTI system (call this $FR(f)$ for now):

for every frequency $f$, the response to $\exp(j2\pi ft)$ is $FR(f)\exp(j2\pi ft)$ which is just a (complex) constant times the input complex exponential.

But $(1)$ shows that $FR(f)$ is just $H(f)$, the Fourier transform of the impulse response $h(t)$, which is what you wanted to prove.