# 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:

 $\ds \norm {T_A \mathbf x}_2^2$ $=$ $\ds \int_0^1 \paren {\int_0^1 \map A {t, \tau} \map {\mathbf x} \tau \rd \tau}^2 \rd t$ 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:

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

$\blacksquare$