Linear Integral Bounded Operator is Continuous

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

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


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

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.

Hence, $T_A$ is a linear transformation.

Furthermore:

Hence:


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

By continuity of linear transformations:


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