Let f be uniformly continuous on (a,b). How do you prove that it is bounded on (a,b)?
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Do you mean an open interval by $(a,b)$ ? – Tony Piccolo Jun 24 '13 at 05:18
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(a,b) is meant to be and open interval – Jun 24 '13 at 05:21
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Do you know the proof in the case of a closed interval ? – Tony Piccolo Jun 24 '13 at 05:24
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HINT: From the definition of uniform continuity (with $\epsilon=1$) you know that there is a $\delta>0$ such that if $x,y\in(a,b)$ and $|x-y|<\delta$, then $|f(x)-f(y)|<1$. For a sufficiently large integer $n$ you can pick points $x_0=a<x_1<x_2<\ldots<x_{n-1}<x_n=b$ so that $x_{k+1}-x_k<\delta$ for $k=0,1,\dots,n-1$. You know that $f$ is bounded on the compact set $[x_1,x_{n-1}]$. If $x\in(x_{n-1},b)$, find an upper bound for $|f(x)|$ in terms of $f(x_{n-1})$; if $x\in(a,x_1)$, find an upper bound for $|f(x)|$ in terms of $|f(x_1)|$. Then put the pieces together to conclude that $f$ is bounded on $(a,b)$.
Brian M. Scott
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Hint: Any uniformly continuous function on a dense set can be extended continuously on the whole set.
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