# Can all linear operators on functions be represented as a convolution of the input function with the operator's impulse response?

Where $f, g$ are functions, $\delta$ is the Dirac Delta distribution ('aka Delta function'), and L is a linear (LTI) operator,

if $g = L(f)$, then can we say in all cases $g = Li * f$ , where $Li$ is defined as $L(\delta)$, i.e. the impulse response?

My questions: Is this true?

Any references would be appreciated.

If $L$ is well-defined linear operator $L^1(\mathbb{R}) \to L^\infty(\mathbb{R})$ then $L[f(y)](x) = h \ast f(x)$ for some $h \in L^\infty(\mathbb{R})$ if and only if $$\forall a, \qquad L[f(y-a)](x) = L[f(y)](x-a)$$ That's why we say "linear and time-invariant (LTI) operators".
The same theoy of LTI operators works in $L^\infty(\mathbb{R}),L^2(\mathbb{R}), C^0(\mathbb{R}), D(\mathbb{R})$ and $D'(\mathbb{R})$. The Dirac delta $\delta$ belongs to this last space : it is a distribution.
• @user45664 Yes $L$ is LTI and if it converges then $h(x) = L[\delta](x) = \lim_{\epsilon \to 0} L[\frac{1_{|y| <\epsilon}}{2 \epsilon}](x)$ – reuns Apr 22 '17 at 18:22
• @user45664 Assume that $f \ast h = 0$ for every $f$ so that $h(x) = 0$ "everywhere". Then $h(0) = 0$ or $h(0) = 1$ doesn't change anything under convolution. So it is unique up to isolated values (sets of measure $0$). – reuns Apr 22 '17 at 18:35
• So you need to state it like this : if $L$ is LTI and $h(x) = \lim_{\epsilon \to 0} L[\frac{1_{|y| <\epsilon}}{2 \epsilon}](x)$ is well-defined, then $L[f] = f \ast h$ @user45664 – reuns Apr 22 '17 at 18:37
• @user45664 Yes, this is a rectangle approaching $\delta$ in the limit, and a rigorous definition of $\delta$. – reuns Apr 22 '17 at 18:51