# Convolution Operator is Continuous Linear Transformation

## Theorem

Let $\map {L^1} \R$ and $\map {L^\infty} \R$ be the real Lebesgue $1$- and real Lebesgue $\infty$-spaces respectively.

Let $f \in \map {L^1} \R$.

Let $f* : \map {L^\infty} \R \to \map {L^\infty} \R$ be the convolution operator such that:

$\ds \forall t \in \R : \forall g \in \map {L^\infty} \R : \map {\paren {f * g} } t = \int_{-\infty}^\infty \map f {t - \tau} \map g \tau \rd \tau$

Then $f*$ is well defined and $f* \in \map {CL} {\map {L^\infty} \R}$, where $CL$ denotes the continuous linear transformation space.

## Proof

 $\ds \forall t \in \R : \forall g \in \map {L^\infty} \R : \ \$ $\ds \size {\map {\paren {f*g} } t}$ $=$ $\ds \size {\int_{-\infty}^\infty \map f {t - \tau} \map g \tau \rd \tau }$ Definition of Convolution Integral $\ds$ $\le$ $\ds \int_{-\infty}^\infty \size {\map f {t - \tau} } \size {\map g \tau} \rd \tau$ $\ds$ $\le$ $\ds \int_{-\infty}^\infty \size {\map f {t - \tau} } \norm g_\infty \rd \tau$ Definition of Supremum Norm $\ds$ $=$ $\ds \norm g_\infty \int_{-\infty}^\infty \size {\map f {t - \tau} } \rd \tau$ $\ds$ $=$ $\ds \norm g_\infty \int_{-\infty}^\infty \size {\map f \sigma } \rd \sigma$ Integration by Substitution, $\sigma = t - \tau$ $\ds$ $=$ $\ds \norm g_\infty \norm f_1$ Definition of P-Seminorm $\ds$ $<$ $\ds \infty$ Definition of Lebesgue Space

Thus, an element of the image of $f*$ is bounded.

Therefore, $f*$ is well-defined.

By definition, $f*$ is integral operator.

By Integral Operator is Linear, $f*$ is linear.

We have that $\norm {f}_1 \in \R$.

$\blacksquare$