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

Question

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.

Linear operator References:
http://mathworld.wolfram.com/LinearOperator.html
http://vergil.chemistry.gatech.edu/notes/quantrev/node14.html
https://www.encyclopediaofmath.org/index.php/Linear_operator

Answer 1

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.