lemma on functions

Question

searching for exercises on continuous function I found one, which I rewrote here :

Def : let $[a;b]$ be a set on which a continuous real function $f$ is defined.Let $[A;B]$ be a subset of $[a;b]$ we say that [A;B] is $f$-dominating over $[a;b]$if

$\forall x \in [a;b], \exists y \in [A;B]$ st. $f(y)\geq f(x)$ (1)

Lemma : let $C \in [A;B]$ then either $[A;C]$ is $f$-dominating over $[a;b]$ or$[C;B]$ is.

Exercise : Prove this lema without using the fact that continuous functions on a close interval are bounded. (ie. without the extreme value theorem)

I started the proof this way

If $[A;C]$ is $f$-dominating then proof is done.

Let's then consider $[A;C]$ is not $f$-dominating. This translates to $$\exists x \in [a;b]\text{ st. } \forall y \in [A;C], f(x)>f(y)$$ But then from $(1)$

...

Thus $[B;C]$ is $f$-dominating.

Can anyone help me filling the dots ?

PS : $f$ is continuous of course ! added to the post

Answer 1

Lemma Let $f$ be a (continuous) function defined on $[a;b]$. Let $[A;B]$ be $f$-dominating over $[a;b]$, and let $C\in[A;B]$ be given. Then either $[A;C]$ or $[C;B]$ is $f$-dominating over $[a;b]$.

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The proof starts off as you stated:

If $[A;C]$ is $f$-dominating, the proof is done.

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Now set $x'\in[a;b]$ such that $$ f(x')>f(y) $$ for all $y\in[A;C]$. Hence $M=f(x')$ is an upper bound on $f([A;C])$. But at the same time there exists $y'\in[A;B]$ such that $$ f(y')\geq f(x')=M $$ Now since $y'\in[A;B]$ but $y'\notin[A;C]$ this implies $y'\in[C;B]$.

(actually it is in (C;B] so it is in [C;B] as well)

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Now let any $x\in[a;b]$ be given:

• If $f(x)\leq M$ we have $f(y')\geq f(x)$ where $y'\in[C;B]$.
• If $f(x)>M$ we cannot find a dominating $f(y)$ over $[A;C]$ and resort to the same argument as for $x',y'$ to show that there exists $y\in[C;B]$ such that $f(y)\geq f(x)$.

Thus $[C;B]$ is $f$-dominating over $[a;b]$.