Let $f:(a,b) \rightarrow \Bbb R$ be continous. Let $g:(a,b) \rightarrow \Bbb R$ be defined by $$g(x):= \sup_{t\in[a,x]}f(t)$$

Show $g$ is also continuous.

Can someone quickly explain how this function is defined? I have trouble understanding it.

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# Proof that $g(x):= \sup_{t\in[a,x]}f(t)$ is continuous

calculusreal-analysis

Let $f:(a,b) \rightarrow \Bbb R$ be continous. Let $g:(a,b) \rightarrow \Bbb R$ be defined by $$g(x):= \sup_{t\in[a,x]}f(t)$$

Show $g$ is also continuous.

Can someone quickly explain how this function is defined? I have trouble understanding it.

## Best Answer

Hints: assuming the OP means $[a,b]$, (because the claim is false as stated (see comment below):

Let $x_0\in [a,b].$

If $\epsilon>0,$ then there is a $\delta>0$ such that $|x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\frac{\epsilon}{2}.$ It follows that if $x_0<x,\ $ and $|x-x_0|<\delta,$

and therefore, using the definition of the supremum,

Similarly, if $x<x_0$ and $|x-x_0|<\delta,$ then $f(x_0)<f(x)+\frac{\epsilon}{2}$ and so