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The image

$d\colon C([a,b])\times C([a,b])\to \mathbb{R}_{\ge 0}\\ (f,g)\mapsto \{\sup |f(t)-g(t)|,t\in [a,b]\}$

defines a metric on the set of all continuous functions. I want to show that

$\Phi_x\colon C([a,b])\to \mathbb{R}\\ f\mapsto f(x)\\ x\in[a,b]$

is continous. I used the absolute value metric on the field of real numbers.

For an arbitrary $\epsilon >0$ I can find a $\delta >0$ that

$|f(x)-g(x)|\le \sup |f(t)-g(t)| <\epsilon$ holds if I choose $\epsilon = \delta$, thus the function is continuous.

Is this correct?

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    Your proof is OK. – 2017-02-04
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    Would there be a more elegant alternative? How would an experienced mathematician do it? – 2017-02-04
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    @EpsilonDelta I gave a slightly expanded version of your proof, as I would have written it as a student. – 2017-02-04

1 Answers 1

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In the sup metric it's quite obvious that for any $x \in [a,b]$: $|\Phi_x(f)| \le \sup\{|f(t): t \in [a,b]\}= ||f||_\infty$, as the sup of a set is an upperbound for it.

As we have a linear map $$|\Phi_x(f) - \Phi(g) | = |\Phi_x(f-g)| \le || f-g||_\infty = d(f,g)$$

So for any $\varepsilon > 0$ and any $g$, we show continuity of $\Phi_x$ at $g$ by picking $\delta = \varepsilon$, then by the above inequality $d(f,g) < \delta$ implies $|\Phi(f) - \Phi(g)| < \delta = \varepsilon$ as well.