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If $f(x)$ is continuous at the closed interval $[a,b]$, and $\forall x_0\in [a,b]$, there is an open interval $U$ containing $x_0$ s.t. $f(x_0)$ is none bigger than $f(\eta)$ for any $\eta\in U$, then prove that $f$ is constant on $[a,b]$.

I think that if we can prove for all $x\in[a,b]$, there is an open interval containing $x$ and $f$ is constant on that interval, then we can use Heine-Borel theorem to prove the statement. But how to prove for all $x$ there is an open interval containing $x$ and $f$ is constant on that interval?

Jiaze Zhang
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Asigan
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    Hint : Let $M$ the upper bound of $f$ on $[a,b]$. There exist $x_0$ such that $f(x_0)=M$. What can you say of $f(x)$ for $x$ close to $x_0$ ? – Kelenner Nov 19 '21 at 16:42
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    @Maksim Well, I want to show that $\forall x_0\in[a,b]$, $\exists(c,d)\subset[a,b]$ s.t. $x_0\in(c,d)$ and $f(\eta)=f(x_0)$, $\forall\eta\in(c,d)$. But how can we show that statement? I tried to assume the contrary. If there is a point $x_0$ s.t. for any open interval $U$ containing $x_0$, there exists $\eta\in U$ s.t. $f(\eta)>f(x_0)$, then this should cause contradiction since for any $x$ in$[a,b]$, there is an open interval containing $x$ and $f$ takes minimal value on $x$ in the interval. But it seems that that method does not work. Could you help me with this method? – Asigan Nov 20 '21 at 03:25

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Following the idea of @Kelenner: As $f$ is continuous on the compact set $[a,b]$ it will attain its maximum $M$ at a certain point $x_{M} \in [a,b]$. Due to the assumption about $f$ there is is an open interval U containing $x_{M}$ such that $f(x_M) \leq f(\eta)$ for all $\eta \in U$. But as $f(x_M)$ is tha maximum this means $$f(x_M) = f(\eta)\qquad \eta \in U$$ This means that $f$ is constant in an open interval and you can try to continue your proof from this point by applying Heine-Borel somehow.

However you can prove the entire statement directly from this reasoning: consider the set $X:=\{x\in[a,b]: f(x) = M\}$ which is nonempty and which is closed (as $f$ is continuous) and as well open. The latter is due to the fact (which we showed above) that for each point, where the maximum is attained there is also an enclosing open interval where $f$ is constantly $M$.

So $X$ is a nonempty clopen subset of $[a,b]$, and as $[a,b]$ is connected, it must hold $X= [a,b]$. Thus $f$ must be constant.

Maksim
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  • Well, sorry for my bothering. But I think the core of my method is to show for all $x_0$, $f$ is constant in a sufficient small open neighbourhood of $x_0$, and I think this is difficult to prove in the general case, in which we do not know $f(x_0)\geq f(x)$ holds for all $x\in[a,b]$ and what we know about $x_0$ is that it is merely an arbitrary point in $[a,b]$. Can you help me prove that statement? – Asigan Nov 21 '21 at 09:30
  • Well at least it can be shown, that for each $x\in[a,b]$ and each non trivial closed interval $I$ containing $x$ there is a sub interval on which $f$ is constant. Maybe you want to start from this? Using the connectivity argument from the answer, you also can conclude that the sub interval must be equal to the interval $I$. – Maksim Nov 21 '21 at 12:45
  • but I got stuck at how to prove that for each $x\in [a,b]$ and each non trivial closed interval I containing x there is a sub interval on which f is constant. Could you help me prove this? – Asigan Nov 21 '21 at 13:21
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    As usual: $f$ will attain its max within this $I$. Just as shown in the answer there is a sub interval of $I$, containing the maximum point (of $f$ on $I$)and due to the local minimum property of $f$ there is a Sub-interval surrounding the maximum point where $f$ is constant. Note that the sub interval does not necessarily contain $x$ – Maksim Nov 21 '21 at 13:32