Riemann Integral Operator is Continuous Linear Transformation

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Theorem

Let $\struct {C \closedint a b, \norm {\, \cdot \,}_\infty}$ be the normed vector space of real-valued functions continuous on $\closedint a b \subseteq \R$ equipped with the supremum norm.

Let $T : C \closedint a b \to \R$ be the Riemann integral operator:

$\ds \forall \mathbf x \in C \closedint a b : \map T {\mathbf x} = \int_a^b \map {\mathbf x} t \rd t$


Then $T$ is a continuous mapping.


Proof

We have that Integral Operator is Linear.

Furthermore:

\(\ds \size {T \mathbf x}\) \(=\) \(\ds \size {\int_a^b \map {\mathbf x} t \rd t}\)
\(\ds \) \(\le\) \(\ds \int_a^b \size {\map {\mathbf x} t} \rd t\)
\(\ds \) \(\le\) \(\ds \int_a^b \norm {\map {\mathbf x} t}_\infty \rd t\) Definition:Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions
\(\ds \) \(=\) \(\ds \paren {b - a} \norm {\mathbf x}_\infty\)

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

$\blacksquare$


Sources