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 $f(a)=f(b)$.

  • $\begingroup$ 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

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 $a$.


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