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Suppose that $f :[a,b]\rightarrow \mathbb{R}$ is continuous. Let $\epsilon>0$. Show that there exists a continuous, piecewise linear function $g: [a,b]\rightarrow \mathbb{R}$ such that $|f(x)-g(x)|<\epsilon$ for all $x$ in $[a,b]$.

Proof: Suppose that $f$ is continuous at $p$. Then for every $\epsilon>0$ there exists a $\delta>0$ such that $ |f(x)-f(p)|<3\epsilon $ for all points $x$ in $[a,b]$ for which $|x-p|<\delta$.

let $|g(p)-f(p)|<\epsilon$ for some $\epsilon$, then

$|f(x)-f(p)|= |f(x)-g(x)+g(x)-g(p)+g(p)-f(p)|\leq|f(x)-g(x)|+|g(x)-g(p)|+|g(p)-f(p)|<3\epsilon$

we get $|g(x)-g(p)|<\epsilon$ since $|f(x)-g(x)|<\epsilon$ and $|g(p)-f(p)|<\epsilon$. Hence there exists a continuous, piecewise linear function $g$.

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3 Answers 3

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Others have already remarked that your $g$ coming out of nowhere is a somewhat fluffy object. There is one more thing: You can do better than just "proving existence". Think of how you would go about constructing such a $g$ when the graph of $f$ is given to you and then make a proof out of your line of thinking.

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Reading your proof, I wonder a few things. Firstly, you say ' let $|g(p) - f(p) | < \epsilon$... does such a $g$ exist? Is it linear? Piecewise linear? Is it even continuous?

These are the problems I have with the proof. But if you update your proof, I'll update this answer.

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I'm regret to say that your proof may not be right.Actually, you have not created continuous, piecewise linear function g(x).My suggestion:1.f is uniformly continuous. 2.You can form the step function at first then work out the continuous one

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