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Question: Let $f$ be a continuous function on $[a,b]$. Show that there is a piecewise linear function $\phi$ on $[a,b]$ with $|f(x)-\phi(x)| < \varepsilon$ for $x \in [a,b]$.

My proof: For any $\varepsilon > 0$ there is a $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y \in [a,b]$. Let $b=a+N\delta$ and partition the interval into subintervals as follows: $x_n=a+n\delta$ for $n=1,2,3,...,N$. Now let $f(x_i)=\phi(x_i)$ and let $\phi(x)$ be piecewise linear between $(x_i,x_{x+1})$.

Then for $x \in (x_i, x_{i+1})$: $$|f(x)-\phi(x)|=|f(x)-f(x_k)+\phi(x_k)-\phi(x)|\leq|f(x)-f(x_k)|+|\phi(x_k)+\phi(x)| < 2\epsilon,$$ since $$|x-x_k|<\delta.$$

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What about $f$? –  Davide Giraudo Sep 17 '12 at 20:16
    
@DavideGiraudo I'm not sure how to incorporate f's uniform continuity, if that is what you are asking. –  emka Sep 17 '12 at 20:18
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No, Davide is pointing out that you never said what kind of function $f$ is. The question should begin with Let $f$ be a continuous real-valued function on $[a,b]$ or words to that effect. –  Brian M. Scott Sep 17 '12 at 20:21
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up vote 2 down vote accepted

Presumably there should be a hypothesis that $f$ is continuous on $[a,b]$. The argument is almost fine. The one slight hiccup is your assumption that $b-a$ is an integral multiple of $\delta$. After you use continuity of $f$ and compactness of $[a,b]$ to get a $\delta$ that ‘works’ for that $\epsilon$, you want to choose $N$ large enough so that $(b-a)/N<\delta$ and then replace $\delta$ by $(b-a)/N$.

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Do you mean that I want to replace the $\delta$ in $a+n\delta$ with $a+n\frac{b-a}{N}$? Also, thanks for noting that I left off a meaningful part of the problem...the continuity of f. –  emka Sep 17 '12 at 20:23
    
@EMKA: Yes. Or use unif. cont. to say that there is a $\delta_0>0$ such that etc., then choose $N$ so that $\frac{b-a}N\le\delta_0$, let $\delta=\frac{b-a}N$, and leave it as is. –  Brian M. Scott Sep 17 '12 at 20:26
    
Continuity $f$ on compact set $[a,\,b]$ implies that $f$ is uniformly continuous . –  M. Strochyk Sep 17 '12 at 20:34
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