Mapping from L1 Space to Real Number Space is Continuous

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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:

\(\ds \forall g \in X: \, \) \(\ds \map {d_1} {f, g}\) \(<\) \(\ds \delta\)
\(\ds \leadsto \ \ \) \(\ds \int_a^b \size {\map f t - \map g t} \rd t\) \(<\) \(\ds \delta\) Definition of $L^1$ Metric
\(\ds \leadsto \ \ \) \(\ds \size {\int_a^b \paren {\map f t - \map g t} \rd t}\) \(<\) \(\ds \delta\) Absolute Value of Definite Integral
\(\ds \leadsto \ \ \) \(\ds \size {\int_a^b \map f t \rd t - \int_a^b \map g t \rd t}\) \(<\) \(\ds \delta\) Linear Combination of Definite Integrals
\(\ds \leadsto \ \ \) \(\ds \size {\map I f - \map I g}\) \(<\) \(\ds \delta\) Definition of $I$
\(\ds \leadsto \ \ \) \(\ds \map d {\map I f, \map I g}\) \(<\) \(\ds \delta\) Definition of $d$
\(\ds \leadsto \ \ \) \(\ds \map d {\map I f, \map I g}\) \(<\) \(\ds \epsilon\) Definition of $\delta$

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.

$\blacksquare$


Sources