Let $f:[0,1]\rightarrow \mathbb{R}$ be a monotonic function with $f\big(\frac14\big)f\big(\frac34\big)\lt 0$ .Suppose
sup$\{x\in [0,1]: f(x)\lt 0\}=\alpha$
Which of the following statements are correct?
$(a)f(\alpha)\lt 0$
$(b)$ if $f$ is increasing, then $f(\alpha)\le 0$
$(c)$ if $f$ is continous and $\frac14 \lt \alpha \lt \frac34$ , then $f(\alpha)=0$
$(d)$ if $f$ is decreasing then $f(\alpha)\lt 0$
My partial soln:
Let $A=\{x\in [0,1] : f(x) \lt 0\}$
For $(c)$
It can't happen that $f(x)$ is monotone decreasing since $f(x_o)=0$ for some $\frac14\lt x_o \lt \frac34$ by IVT which implies $f(x)\lt 0,\forall x\gt x_o$ and thus $\alpha\gt \frac34$, a contradiction.
$\alpha\in \overline {A}\subset \overline{[0,1]}=[0,1]$
If possible,Let $f(\alpha)\gt 0$ .Then by continuity of $f$ at $\alpha, \exists \epsilon \gt 0$ such that $\forall x\in N(\alpha, \epsilon)$,we have $f(x)\gt 0$.
Now let $y\in A$ be arbitary, and $x_o\in (\alpha-\epsilon,\alpha)$ then $y\lt x_o$ otherwise $ 0 \gt f(y) \ge f(x_o) \gt 0$, thus contradiction. Since $y$ was arbitary and $x_o\lt \alpha$, this contradicts that $\alpha$ is the supremum.Hence $f(\alpha)\le 0$.
Similar reasoning will allow us to show $f(\alpha)\lt 0$ is not possible, so $f(\alpha)=0$.
$(c)$ ends...
For $(d)$
We must have $f(\frac34)\lt 0$ and since $f$ is decreasing $f(x) \lt 0, \forall x\gt \frac34$ and thus $\alpha=1$ implying $f(\alpha)\le f(\frac 34)\lt 0$
I am confused with $(a)$ and $(b)$ and I humbly request you to check if my arguments for $(c)$ and $(d)$ are correct.
Thanks for your valuable time!!
Your arguments for c) and d) are OK. a) and b) are both false and there is a simple counter-example: let $f(x)=-1 $ for $0 \leq x <\frac 1 2$ and $f(x)=1$ for $\frac 1 2 \leq x \leq 1$. Note that $\alpha=\frac 1 2$ and $f(\alpha) >0$.