If $f : [0,1] \longrightarrow \Bbb R$ be a monotonic function with $f \left (\frac {1} {4} \right ) f \left (\frac {3} {4} \right ) \lt 0.$ Let $\sup \{x \in [0,1]\ \big |\ f(x) \lt 0 \} = \alpha.$ Then show that
$(1)$ If $f$ is continuous and $\frac {1} {4} \lt \alpha \lt \frac {3} {4},$ then $f(\alpha) = 0.$
$(2)$ If $f$ is decreasing then $f(\alpha) \lt 0.$
My attempt $:$ $(1)$ If $f$ is continuous with $\frac {1} {4} \lt \alpha \lt \frac {3} {4},$ then $f$ can't be decreasing because if $f$ were decreasing then the relation $f \left (\frac {1} {4} \right ) f \left (\frac {3} {4} \right ) \lt 0$ yields $f \left (\frac {1} {4} \right ) \gt 0$ and $f \left (\frac {3} {4} \right ) \lt 0.$ But then for all $\frac {3} {4} \lt x \leq 1$ we have $f(x) \leq f \left (\frac {3} {4} \right ) \lt 0.$ Hence $\alpha = 1,$ a contradiction to the fact that $\frac {1} {4} \lt \alpha \lt \frac {3} {4}.$ This shows $f$ has to be increasing. So $f (x) \geq 0,$ for all $x \gt \alpha.$ Since $f$ is continuous at $\alpha,$ letting $x$ tending to $\alpha$ from the right it follows that $$f(\alpha) \geq 0\ \ \ \ \ \ \ \ (1)$$
Now if $f(\alpha) \gt 0$ then since $f$ is continuous on $[0,1]$ and hence in particular on $\left [\frac {1} {4}, \alpha \right ]$ with $f \left (\frac {1} {4} \right ) \lt 0$ it follows from intermediate value theorem that there exists $x_0 \in \left (\frac {1} {4}, \alpha \right )$ such that $f(x_0) = 0.$ But since $f$ is increasing it follows that for all $x \in [x_0, 1]$ we have $f(x) \geq 0.$ But that means $\alpha \leq x_0,$ a contradiction to the fact that $x_0 \in \left (\frac {1} {4}, \alpha \right ).$ So $f (\alpha ) \not \gt 0$ i.e. $$f(\alpha) \leq 0\ \ \ \ \ \ \ \ (2)$$ Combining $(1)$ and $(2)$ we have $f(\alpha) = 0,$ as required.
$(2)$ Since $f$ is decreasing it follows from the given relation $f \left (\frac {1} {4} \right ) f \left (\frac {3} {4} \right ) \lt 0$ that $f \left (\frac {1} {4} \right ) \gt 0$ and $f \left (\frac {3} {4} \right ) \lt 0.$ But then for all $\frac {3} {4} \lt x \leq 1$ we have $f(x) \leq f \left (\frac {3} {4} \right ) \lt 0.$ This shows that $\alpha = 1$ and hence $f(\alpha) = f(1) \lt 0,$ as required.
This completes the proof.
QED
Can anybody please check my solution above whether it holds good or not? Thanks for your time.