Mapping from L1 Space to Real Number Space is Continuous

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Theorem

Let $\left({\R, d}\right)$ be the real number line under the usual metric $d$.

Let $X$ be the set of continuous real functions $f: \left[{a \,.\,.\, b}\right] \to \R$.

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

Let $I: X \to \R$ be the real-valued function defined as:

$\displaystyle \forall f \in X: I \left({f}\right) := \int_a^b f \left({t}\right) \ \mathop d t$


Then the mapping:

$I : \left({X, d_1}\right) \to \left({\R, d}\right)$

is continuous.


Proof

The $L^1$ metric on $X$ is defined as:

$\displaystyle \forall f, g \in S: d_1 \left({f, g}\right) := \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t$

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

Let $f \in X$.

Let $\delta = \epsilon$.

Then:

\(\, \displaystyle \forall g \in X: \, \) \(\displaystyle d_1 \left({f, g}\right)\) \(<\) \(\displaystyle \delta\)
\(\displaystyle \implies \ \ \) \(\displaystyle \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t\) \(<\) \(\displaystyle \delta\) Definition of $L^1$ Metric
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\int_a^b \left({f \left({t}\right) - g \left({t}\right)}\right) \ \mathrm d t}\right\vert\) \(<\) \(\displaystyle \delta\) Absolute Value of Definite Integral
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\int_a^b f \left({t}\right) \ \mathrm d t - \int_a^b g \left({t}\right) \ \mathrm d t}\right\vert\) \(<\) \(\displaystyle \delta\) Linear Combination of Integrals
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{I \left({f}\right) - I \left({g}\right)}\right\vert\) \(<\) \(\displaystyle \delta\) Definition of $I$
\(\displaystyle \implies \ \ \) \(\displaystyle d \left({I \left({f}\right), I \left({g}\right)}\right)\) \(<\) \(\displaystyle \delta\) Definition of $d$
\(\displaystyle \implies \ \ \) \(\displaystyle d \left({I \left({f}\right), I \left({g}\right)}\right)\) \(<\) \(\displaystyle \epsilon\) Definition of $\delta$

Thus it has been demonstrated that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall g \in X: d_1 \left({f, g}\right) < \delta \implies d \left({I \left({f}\right), I \left({g}\right)}\right) < \epsilon$

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