Prove that the function $I(f)=\int_{-1}^1 f(x)\, dx$ is continuous on the metric space: $(C([-1,1],\mathbb R), d)$, where $d(f,g) = \sup_{x \in [-1,1]}|f(x) - g(x)|$
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Sorry for the bad formatting, my first time :P – user501050 Nov 09 '17 at 12:09
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1Welcome to MSE. What did you try? – José Carlos Santos Nov 09 '17 at 12:12
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Thanks for the edit :), I'm having a hard time understanding how it is continuous given such weird distance metric. – user501050 Nov 09 '17 at 12:15
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What does it mean to say that $I$ is continuous? – José Carlos Santos Nov 09 '17 at 12:16
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1it may be useful to know that $\int_a^b f(x) dx \leq \sup_{[a,b]}f(x)\cdot \lvert b -a \rvert$. – ziggurism Nov 09 '17 at 12:16
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I is continuous on $g \in C$ if there exists $\epsilon>0$ and $\delta>0$ such that $d(I(f),I(g)) < \epsilon$ whenever $d(f,g) < \delta$. – user501050 Nov 09 '17 at 12:27
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@user501050 your definition of continuity at $g$ is not correct.. $I$ is continuous at $g\in C([0,1], \mathbb{R})$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $d(I(f), I(g))<\varepsilon$ whenever $d(f,g)<\delta$ – bradipolpo Nov 13 '17 at 10:46