Linear Integral Bounded Operator is Continuous

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Theorem

Let $I = \closedint 0 1$ be a closed real interval.

Let $A : I \times I \to \R$ be a real function such that:

$\ds \int_0^1 \int_0^1 \paren {\map A {t, \tau} }^2 \rd t \rd \tau < \infty$


Further research is required in order to fill out the details.
In particular: For now "bounded" means above. Need to check if this meaning is standard
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where $\times$ denotes the cartesian product.

Let $T_A : \map {L^2} I \to \map {L^2} I$ be an integral operator such that:

$\ds \map {\paren {T_A \mathbf x} } t := \int_0^1 \map A {t, \tau} \map {\mathbf x} \tau \rd \tau$

where $\mathbf x \in \map {L^2} I$, and $\map {L^2} I$ is the Lebesgue $2$-space.


Then $T_A$ is a continuous transformation.


Proof

We have that Riemann Integral Operator is Linear Mapping.


Further research is required in order to fill out the details.
In particular: Probably this should be replaced with Lebesgue integral. The source does not say anything about compatibility of Riemann integral and Lebesgue space
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Hence, $T_A$ is a linear transformation.

Furthermore:

\(\ds \norm {T_A \mathbf x}_2^2\) \(=\) \(\ds \int_0^1 \paren {\int_0^1 \map A {t, \tau} \map {\mathbf x} \tau \rd \tau}^2 \rd t\) Definition of P-Seminorm
\(\ds \) \(\le\) \(\ds \int_0^1 \paren{\int_0^1 \paren{\map A {t, \tau} }^2 \rd \tau} \paren {\int_0^1 \paren{\map {\mathbf x} \tau}^2 \rd \tau }\rd t\) Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals
\(\ds \) \(\le\) \(\ds \paren {\int_0^1 \int_0^1 \paren{\map A {t, \tau} }^2 \rd \tau \rd t} \paren {\int_0^1 \paren{\map {\mathbf x} \tau}^2 \rd \tau }\)
\(\ds \) \(=\) \(\ds \paren {\int_0^1 \int_0^1 \paren{\map A {t, \tau} }^2 \rd \tau \rd t} \norm {\mathbf x}_2^2\) Definition of P-Seminorm
\(\ds \) \(<\) \(\ds \infty\)

Hence:

$T_A \mathbf x \in \map {L^2} I$.

By continuity of linear transformations:

$T_A \in \map {CL} {\map {L^2} I}$.

$\blacksquare$

Sources

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