Jtdylktuy

Let $f:[a,b] to Bbb R$ be continuous . Define $g:[a,b] to Bbb R$ by $$g(x) = sup[f(t) | t in [a,x]].$$

Then is $g$ continuous? I tried to show the inequality $$|g(x)-g(y)| le sup(f(t) | t in [x,y])$$ but wasn’t able to. Any hints or suggestions will be appreciated.

Let $aleqslant xlt y$. Then
$$
g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
$$
We use the bound
$$
sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
$$
in order to get
$$
g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
$$
and since $g$ is non-decreasing, it follows that
$$
leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
$$
Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.

Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by

$$f(x) = 2x-1.$$

then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?

For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.

Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?