1
$\begingroup$

Here's the proof outline by Spivak: Here's the proof outline by Spivak

I don't understand why do we need part a) (I can prove it though). Why do the following proof won't work?

Suppose $f(a)>f(b)$.
Since $f$ is continuous on $R$, it is continuous at $a$, and so $\exists\delta>0\ \forall x \ |x-a|<\delta\implies f(x)>f(b)$.
So, we have $x\in(a,b)$, and $f(x)>f(b)$, which contradicts the statement that all points in $(a,b)$ are "shadow points".
Which leaves us with only possibility that $f(a)=f(b)$.

$\endgroup$
  • $\begingroup$ Well, technically speaking in order to reach a contradiction in your argument above you should also conclude that there is no point $y$ in $(a,b)$ such that $f(y)>f(b)$. Doing this amounts to proving that $f$ attains its maximum at $a$. $\endgroup$ – Leo163 Jan 14 at 13:04
  • $\begingroup$ @Leo163 Yes, but isn't it already stated in the problem, "... all points of $(a,b)$ are shadow points..." (which is equivalent to $x\in(a,b) \implies \exists y>x$ with $f(y)>f(x)$)? $\endgroup$ – yellowcat Jan 14 at 13:26
  • $\begingroup$ That implication is not immediate, at least to me: it could be that for every point $y\in (a,b)$ there is a point $z\in (y,b)$ such that $f(z)>f(y)$. The fact that this is not the case is something that requires a proof, at least in my perspective. $\endgroup$ – Leo163 Jan 14 at 14:39
0
$\begingroup$

Your proof is correct. It boils down to the fact that if $f(a) > f(b)$, continuity at $a$ would mean $f$ is still be $> f(b)$ in a neighborhood of $a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.