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

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?

  • 2
    $\begingroup$ There is no assumption that $f$ is continuous in the statement. So we can easily come up with counterexamples. $\endgroup$ Mar 28 '20 at 18:25
  • $\begingroup$ 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. $\endgroup$
    – alortimor
    Mar 29 '20 at 13:45
  • $\begingroup$ 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 $\endgroup$ Mar 29 '20 at 16:29

False, since $f$ is not assumed continuous. For a concrete counterexample, consider

$$f: [-1,1] \to \mathbb{R}: x \mapsto \begin{cases}1/x \quad x \neq 0 \\ 0 \quad \quad x =0\end{cases}$$

A maximum nor a minimum is attained, as this function is unbounded.


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