Convolution Operator is Continuous Linear Transformation

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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.


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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.

$\blacksquare$


Sources

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