Vtyjkyuk

Let $f:[a,b]toBbbR$ be continuous. Does beginalignmaxf(x) endalign exist?

MY WORK

I believe it does and I want to prove it.

Since $f:[a,b]toBbbR$ is continuous, then $f$ is uniformly continuous. Let $epsilon> 0$ be given, then $exists, delta>$ such that $forall x,yin [a,b]$ with $|x-y|<delta,$ it implies $|f(x)-f(y)|<epsilon.$

Then, for $aleq xleq b,$

beginalign f(x)=f(b)+[f(x)-f(b)]endalign
beginalign |f(x)|leq |f(b)|+|f(x)-f(b)|endalign
beginalign maxlimits_aleq xleq b|f(x)|leq |f(b)|+maxlimits_aleq xleq b|f(x)-f(b)|endalign

I am stuck at this point. Please, can anyone show me how to continue from here?

Because $[a,b]$ is compact, every sequence in $[a,b]$ has a subsequence that limits to a point in $[a,b]$. Pick a sequence $x_n in [a,b]$ such that $lim_n rightarrow infty |f(x_n)| = sup_[a,b]|f|$. Now get a convergent subsequence $x_n_k$ that converges to some $x in [a,b]$.

By continuity of $|f|$,

$|f(x)| = |f( lim_krightarrow inftyx_n_k)| = lim_krightarrow infty |f(x_n_k)| = lim_n rightarrow infty |f(x_n)| = sup_[a,b]|f|$.

Here's what I would do, in a proof sketch.

Claim: If $f:[a,b] rightarrow mathbbR$ is continuous, then $f([a,b])$ has a maximal element.

Proof Sketch: Since the set $[a,b] subset mathbbR$ is closed and bounded, by the Heine-Borel (or Bolzano-Weierstrass) Theorem it is compact. Since $f:[a,b] rightarrow mathbbR$ is continuous, it preserves compactness, so $f([a,b]) subset mathbbR$ is also compact. But since $f([a,b]) subseteq mathbbR$ is compact, again by Heine-Borel it is closed and bounded. Thus, we can construct a sequence of points converging to the supremum of $f([a,b])$, and since $f([a,b])$ is a closed set it must also contain the limit, which is precisely the maximal element of $f([a,b])$ we wanted.

If you have any questions about any of the arguments used in the Proof Sketch, do not hesitate to ask.

Regards!

-Dan

you can prove in an another way

as $f:[a,,b] to Bbb R$ so $f([a,b])=[m,M]$ if $f(x) ge 0$ then its obvious that $beginalignmaxf(x) endalign$ exists. similarly if $f(x) le 0$ then $f([a,b])=[-m,-M]$ where $m,M ge 0$ then again $beginalignmaxf(x) endalign$ exists and $beginalignmaxf(x) endalign=m$.

in similar manner u can prove for when $f([a,b])=[-m,M]$ where $m,M ge 0$