Continuity

**mms** · August 16th 2009, 01:09 PM

Let f be a continous function in [a,b] and
$ g(x) = \left\{ \begin{gathered} f(a)\,\,\,x = a \hfill \\ \max \left\{ {f(x):\,x \in \left[ {a,x} \right]} \right\}\,\,\,x \in (a,b] \hfill \\ \end{gathered} \right. $

prove that g is continous in [a,b]

**mms** · August 16th 2009, 01:24 PM

hmm well the problem says that g(x)= max f(x) where x is in [a,x] :/

**arbolis** · August 16th 2009, 01:28 PM

Originally Posted by mms

hmm well the problem says that g(x)= max f(x) where x is in [a,x] :/

Yes sorry, I misread this part.

**ynj** · August 16th 2009, 05:09 PM

Originally Posted by mms

Let f be a continous function in [a,b] and
$ g(x) = \left\{ \begin{gathered} f(a)\,\,\,x = a \hfill \\ \max \left\{ {f(x):\,x \in \left[ {a,x} \right]} \right\}\,\,\,x \in (a,b] \hfill \\ \end{gathered} \right. $

prove that g is continous in [a,b]

$\forall\epsilon>0,\exists|\delta|>0,\forall|h|<\de lta,|f(x+h)-f(x)|<\epsilon.$
since $g(x+h)=\max\{g(x-|h|),max\{f(x)|x\in[x-|h|,x+|h|\}\}$ , so if $g(x+h)=f(y)$ for some $y\in[a,x-|h|]$ , then $g(x+h)-g(x)=0$ ,if $g(x+h)=f(y)$ for some $y\in[x-|h|,x+|h|]$ ,then $|g(x+h)-g(x)|=|f(y)-g(x)|\leq|f(y)-f(x)|\leq\epsilon$ .In any case, $|g(x+h)-g(x)|\leq\epsilon$ . So $g(x)\in C[a,b]$ .

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