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Let $R$ be the real line with the standard metric $d:R \times R \to R$ be defined by $d(x,y) = |x-y|$.

Let $X$ be the set of continuous functions $f:[a,b] \to R$ of an arbitrary closed interval $[a,b]$. In other words, let $X = C[a,b]$ (in some notations).

Let $\rho: X \times X \to R$ be the metric on the functions $f,g \in X$, where $$\rho(f,g) = \int_a^b \left|f(t) - g(t)\right|\;dt.$$

Let $I:(X,\rho) \to (R,d)$ be defined by $$I(f) = \int_a^b f(t)\;dt.$$ We know that $I$ is well-defined because you can integrate continuous functions over compact sets.

Prove that the function $I$ is continuous.

I can only get to a certain point in proving this before I get stuck.

Let $\epsilon > 0$ be arbitrary. Suppose $\delta > 0$. Then we want to show that if $\rho(f,g) < \delta$, then $d(I(f), I(g)) < \epsilon$.

The first inequality is expanded to $$\int_a^b\left|f(t)-g(t)\right|\;dt < \delta.$$

The second inequality is expanded to $$\left|\int_a^bf(t)\;dt - \int_a^bg(t)\;dt\right| < \epsilon,$$ or $$\left|\int_a^b[f(t) - g(t)]\;dt\right| < \epsilon.$$

It is at this point that I'm stuck. When I originally attempted this proof, I conjectured that $\left|\int_a^bf(t)\;dt\right| = \int_a^b\left|f(t)\right|\;dt$ so that we could simply set $\delta = \epsilon$, but I don't believe this is true in general. I'd imagine some variant of the Cauchy-Shwarz inequality might help, but I'm not sure how exactly.

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By integrating $-|f(t)|\le f(t)\le |f(t)|$, you can obtain $|\int_a^bf(t)dt|\le \int_a^b|f(t)|dt$.

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Perfect! Then we can let $\delta = \epsilon$, and the continuity follows. –  Matt Nov 14 '12 at 20:57
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