Let $f:[a ,b)\to \mathbb{R}$ be a continous function. Define $F:[a,b)\to \mathbb{R}$ by
$F(x)=\displaystyle\max_{t \in [a,x]}f(t)$
Then which of the following are true?
$(1)F$ need not be continous.
$(2)F$ is necessarily monotonic.
$(3)F$ is necessarily bounded.
$(4)F$ is Riemann Integrable on $[a,x]$ for every $x\in [a,b)$
My attempt: -
F(x) is monotone.
Let $x_2 \gt x_1$ from $[a ,b)$
Then $F(x_2)=\displaystyle\max_{t\in [a,x_2] }f(t) \ge \displaystyle\max_{t\in [a,x_1]}f(t)=F(x_1)$
To prove $F$ is continous.
Let $c \in [a,b)$
My plan is to show that for sequences approaching to $c$ from left and right , the correspoding functional sequence with repsect to $F$ converges to $F(c)$
Let $\{x_n\}$ be a sequence such that $x_n \lt c$ and $x_n \to c$ i.e it approaches $c$ from the left.
Let $\{y_n\}$ be a sequence such that $y_n \gt c$ and $y_n \to c$
Now since $f$ is continous , max of $f$ on $[a,c]$ is attained there.
Case $(1)$
$f(c) \lt \displaystyle\max_{t\in [a,c] }f(t)=f(\alpha) $ (say), where $a\le \alpha \lt c$
Then $F(c)=f(\alpha)$
To show $\{F(x_n)\}$ converges to $F(c)$
Then for $\alpha \le x \le c$, we have $F(x)=f(\alpha)$, a constant function .
So for a large $n$ , we have $\{F(x_n)\}$ as the constant sequence $\{f(\alpha)\}$ and so
$F(x_n)=f(\alpha) \to f(\alpha)=F(c)$ as $n\to \infty$
To show $\{F(y_n)\}$ converges to $F(c)$
Let $\epsilon=\frac{f(\alpha)-f(c)}2$
Then for this $\epsilon , \exists \delta \gt 0$ such that $|f(c)-f(x)|\lt \epsilon , \forall x\in N(c ,\delta)$
So for $c\le x \le c+\delta$, we have $f(x)\lt f(c)+\epsilon =\frac{ f(\alpha)+f(c)}2 \lt f(\alpha)$
We then have $\displaystyle\max_{t\in[a,c+\delta]} f(t)=f(\alpha)$
Then for a large enough $n$ , we have $\{F(y_n)\}$ as the constant sequence $\{f(\alpha)\}$ and so it converges to $f(\alpha)=F(c)$
Case $1$ finishes here,
Case $2$)
$\displaystyle\max_{t \in [a ,c] } f(t) =f(c)$ or $F(c)=f(c)$
Since $f$ is continous at $c$, so for $\epsilon' \gt 0, \exists m\in \mathbb{N}$ such that $\forall n \gt m$, we have
$|f(x_n)-f(c) |\lt \epsilon' \quad (*)$
$\Rightarrow f(c)-\epsilon' \lt f(x_n ) \lt f(c)+\epsilon'$
To show $\{F(x_n)\}$ converges to $F(c)$
Now $F(c) \ge F(x_n ) \ge f(x_n) \gt f(c) -\epsilon'$ for $n\gt m$
But $F(c)=f(c)$
So for $n\gt m$, we have
$F(c)-\epsilon' \lt F(x_n) \le F(c)$
Thus giving $\displaystyle\lim_{n\to \infty}F(x_n)=F(c)$
To show $\{F(y_n)\}$ converges to $F(c)$
From $(*)$ and combining the fact $F(c)=f(c)$ it follows that $F(y_n) \lt f(c)+\epsilon'=F(c)+\epsilon'$ for a large enough $n$ , say for all $n\gt M$
So for all $n\gt M$, we have
$F(c) \le F(y_n) \lt F(c)+\epsilon'$ and thus showing $\displaystyle\lim_{n\to \infty}F(y_n)=F(c)$ .
Case $(2)$ ends here. and $F$ is shown continous.
$(4)$ is obviously true.
Are my works correct? Any way to shorten it?
Will be glad to know other ideas. Thanks for your time.