Two statements about one-sided derivative and monotony

Question

The statement 1 is: $f\colon [a,b]\to\mathbb R$,continuous on $[a,b]$,$f'_-(x)$ exists and is $\le0$ for all $(a,b]$.Can we infer that f is non-increasing on $[a,b]$?

My attempt is: Assume $f$ is not non-increasing.Then there exists $a_1\lt b_1$ and $f(a_1)\lt f(b_1)$.Let $c=inf\{x\in[a_1,b_1]\big|f(x)=f(b_1)\}$.Since $f$ is continuous,we can know that:1.$f(c)=f(b_1)$,and 2.there is no $x\in[a_1,c)$ such that $f(x)\ge f(b_1)$.$\ \ $(For if there exists such $x$,by the definition of c,$f(x)\neq f(b_1)$,so $f(x)\gt f(b_1)\gt f(a_1)$,then by intermediate value theorem,there exists $t\in (a_1,x)$ and $f(t)=f(b_1)$,which violates the definition of c.) It follows that for all $h<0$ which satisfies $(c+h)\ge a_1$,we have $\frac{f(c+h)-f(c)}{h}\gt 0$.Because the left-sided derivative exists,we know the left-sided limit of this inequility exists.Taking the limit,we have $f'_-(c)\ge 0$,which I think is not enough to lead to a contradiction.

Besides,in Continuous function with continuous one-sided derivative, there is a slightly different statement in the third line of the answer.Let's call it statement 2.

My questions:
1.Is the statement 1 true or false?
2.I think my attempt can be used to prove the statement 2,but I'm not sure,because the two statements look quite similar,and it fails for statement 1. So,is my attempt true for statement 2 and if statement 1 is true can it be improved for proving the statement 1?
3.If statement 1 is true,can the method by the answerer in the above link be adapted to prove statement 1? I haven't fully understood it now because I'm new to the transfinite induction.

Any help/advice is appreciated.

Answer 1

The conclusion you draw from $f'_-(c) \ge 0$ doesn't contradict your hypothesis; it just implies that $f'_-(c) = 0$.

On the other hand, suppose that $f'_-(x) < 0$ (strict inequality) for all $x \in (a,b]$. Then you can conclude that $f$ is strictly decreasing. Otherwise there would be points $c,d \in [a,b]$ with $c < d$ and $f(c) \le f(d)$. Since $f$ is continuous it possesses a maximum on the interval $[c,b]$ at a point $x_0 \not= c$. But this implies $f'_-(x_0) \ge 0$, which is a contradiction.

This can be used to prove the Lemma used in statement 2 with no reference to transfinite induction. How can you get from here to your statement?

Define, for each $n$, $f_n(x) = f(x) - \frac xn$. If $f'_-(x) \le 0$ for all $x \in (a,b]$, then $(f_n)'_-(x) \le - \frac 1n$ for all $x \in (a,b]$. Thus each $f_n$ is strictly decreasing, and $f$ (being a uniform limit of decreasing functions on $[a,b]$) is thus nonincreasing.