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
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 $\norm f_1 \in \R$.

By Continuity of Linear Transformation between Normed Vector Spaces, $f*$ is continuous.