Mapping from L1 Space to Real Number Space is Continuous

Theorem
Let $\struct {\R, d}$ be the real number line under the usual metric $d$.

Let $X$ be the set of continuous real functions $f: \closedint a b \to \R$.

Let $d_1$ be the $L^1$ metric on $X$.

Let $I: X \to \R$ be the real-valued function defined as:
 * $\ds \forall f \in X: \map I f := \int_a^b \map f t \ \mathop d t$

Then the mapping:
 * $I: \struct {X, d_1} \to \struct {\R, d}$

is continuous.

Proof
The $L^1$ metric on $X$ is defined as:
 * $\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$

Let $\epsilon \in \R_{>0}$.

Let $f \in X$.

Let $\delta = \epsilon$.

Then:

Thus it has been demonstrated that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall g \in X: \map {d_1} {f, g} < \delta \implies \map d {\map I f, \map I g} < \epsilon$

Hence by definition of continuity at a point, $I$ is continuous at $f$.

As $f$ is chosen arbitrarily, it follows that $I$ is continuous for all $f \in X$.

The result follows by definition of continuous mapping.