# True or False: let $f:[a,b] \rightarrow \Bbb{R} \; \exists \, c \in [a,b] \text{ s.t. } f(c) \geq f(x) \; \forall x \in [a,b]$

$$f:[a,b] \rightarrow \Bbb{R} \; \exists \, c \in [a,b] \text{ s.t. } f(c) \geq f(x) \; \forall x \in [a,b]$$

So this is what I've come up with:

True, since if $$\exists \, c \in [a,b] \text{ s.t. } f(c) \geq f(x) \; \forall x \in [a,b]$$, then let $$s= \sup{f}$$

Need to show that $$s$$ is absolute maximum of $$f$$.

Absolute maximum must exist in $$[a,b]$$, whereas $$\sup{f}$$ need not, in which case, $$\exists \, c \in [a,b] \text{ s.t. } f(c) \geq f(x) \, \forall \, x \in [a,b]$$ would be true, as $$c$$ would represent the absolute maximum, therefore

$$\exists \, x_{1} \in [a,b] \text{ s.t. } s-\frac{1}{n} \lt f(x_{1}) \leq s \text{ *}$$

Every bounded sequence converges to a number, therefore let $$x_{2}$$ be such a number that $$[a,b]$$ converges to $$\Rightarrow \, x_{2} \in [a,b]$$

$$s-\frac{1}{n} \lt f(x_{2}) \leq s \text{ **}$$

At this point, I am unsure how to employ the IVT by using * and ** to prove that $$f$$ therefore has an absolute maximum and it is $$s$$?

Alternatively, is there perhaps a simpler way to prove true or false?

• There is no assumption that $f$ is continuous in the statement. So we can easily come up with counterexamples. Mar 28 '20 at 18:25
• What if $f$ was defined as $\text{let }f:[a,b] \rightarrow \Bbb{R} \text{ be bounded } \; \exists \, c \in [a,b] \text{ s.t. } f(c) \geq f(x) \; \forall x \in [a,b]$, so $f$ is defined as bounded, but not continuous. Mar 29 '20 at 13:45
• Bounded still isn't enough. You can still construct counterexamples. Consider a continuous function on $[a,b]$. Modify it by creating a "hole" where the function has a maximum Mar 29 '20 at 16:29

False, since $$f$$ is not assumed continuous. For a concrete counterexample, consider
$$f: [-1,1] \to \mathbb{R}: x \mapsto \begin{cases}1/x \quad x \neq 0 \\ 0 \quad \quad x =0\end{cases}$$