1. ## Continuity

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

prove that g is continous in [a,b]

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

3. 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.

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

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