$f: [a,b]\to\Bbb R $. If $g(x)=\sup\{f(t):t∈[a,x]\}$ and $f$ is continuous , then prove that $g$ is continuous at $a$.
My answer :
Because $f$ is continuous in $[a,b], f$ is continuous at $a$. So, you have that $$\displaystyle\lim_{t\to a} f(t)=f(a)$$
But $g(a)=\sup\{f(t):t\in[a,a]\}=\sup\{f(a)\}=f(a)$ $$\therefore g(a)=f(a)$$
Therefore, $\displaystyle\lim_{x\to a} g(x)=g(a)=f(a)$. So with this I know that
1) $\displaystyle\lim_{t\to a}f(t)=f(a)$ so $f$ is continuous at a ?
2) and it is $\displaystyle\lim_{t\to a}f(t)$ or $\displaystyle\lim_{x\to a}f(x)$?
Since $f(t)$ is a continuous continous function over a closed interval, $f(t)$ achieves its maximum.
$g(x)$ is a monotonically increasing function.
For all $x_0$ in $[a,b)$
With $x > x_0$, either $f(t)$ achieves a new maximum in $[x_0, x]$ i.e. $\sup \{f(t): t\in [x_0,x]\} > \{f(t): t\in [a,x_0]\}$ in which case
$g(x) - g(x_0) < \sup \{f(t) - f(x_0): t\in[x_0,x]\}$
or it doesn't and
$g(x) - g(x_0) = 0$
In either case $|g(x) - g(x_0)| < \{f(t) - f(x_0): t\in[x_0,x]\}$
And, since f(t) is continuous, for any $\epsilon >0$ there is a $\delta > 0$ such that $|x-x_0|<\delta \implies |g(x) - g(x_0)|\le |f(x) - f(x_0)| < \epsilon$
and $g(x)$ is continuous