Equivalence of EVT consequence

Question

I was getting over EVT whics states that if a real-valued function $f$ is continuous on the closed interval $[a,b]$ then $f$ must attain a maximum and a minimum, each at least once. Then there is one consequence which states following: If $f:[a,b]\rightarrow\mathbb{R}$ is continuous $\implies$ $f([a,b])=[f(x_{max}),f(x_{min})]$. Also there is some theorem which states that if $f$ is monotonic, then $\Longleftarrow$ holds.

My question is, why doesn't $\Longleftarrow$ hold if $f$ is not monotonic?

I would be very grateful if someone would explain this to me.

Answer 1

Take $$[a,b]=[0,4\pi]$$

$$f(x)=\sin(x)\;\; if \;\; x\ne \frac{\pi}{4}$$ $$f(\frac{\pi}{4})=\frac 12$$ $f$ is not continuous at $[a,b]$.

$f$ is not monotonic.

$$f([a,b])=[-1,1]=[f(\frac{3\pi}{2}),f(\frac{\pi}{2})]=[\min f,\max f]$$

Answer 2

You can build counter examples with functions that oscillate very quickly near some points, for instance $f(x) = sin(\frac{1}{x})$ for $x \neq 0$ and $f(0)=0$. This map is not continuous at point $0$, but satisfies the property that the image of each interval is an interval.