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I have the following definition of Local boundedness:

function $f(x)$ in $I\subseteq R$ called local bounded if for every $x_0\in I$ exists $\delta>0$ so $f(x)$ is bounded in $(x_0-\delta,x_0+\delta)$.

I was asked to prove that if function is defined in closed interval and locally bounded then it is bounded.

From previous threads I saw that there is a similar theorem for continuous function. I was asked to prove for a defined function. Is it a mistake? Is the proof valid without the continuous condition? if so, how to prove it?

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  • $\begingroup$ We can show that a continuous $f:I\to \Bbb R$ is locally bounded if $I$ is an interval in $\Bbb R$. Now apply the more general result of your Q to conclude that if $I$ is a closed bounded real interval and $f:I\to \Bbb R$ is continuous then $f$ is bounded. $\endgroup$ Commented Jan 4, 2020 at 1:49

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I will assume that the domain is a closed and bounded interval $[a,b]$. Otherwise $f(x)=x$ in $\mathbb R$ would be a counter-example.

Continuity is not required. For each $x_0$ there is an open interval $I_{x_0}$ around it on which $f$ is bounded. By compactness of $[a,b]$ there are a finite number of these intervals, say $I_1,I_2,...,I_n$ that cover $[a,b]$. If $|f| \leq M_j$ in $I_j$ and $M=\max \{M_1,M_2,...,M_n\}$ the $|f(x)| \leq M$ for all $x$.

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    $\begingroup$ But there are closed and unbounded intervals. For those the assertion ist not true as vor instance $f$ with $f(x):=x$ is a counterexample, $\endgroup$ Commented Jan 4, 2020 at 6:02
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May I assume that your interval is bounded? If yes, the point is you can cover any bounded interval $[a,b]$ by a finite number of open interval of the form $$(a_{i} - \delta, a_{i} + \delta)$$ by doing this you use your hypothesis on each $$ (a_{i} - \delta, a_{i} + \delta).$$ A similar result can be proved for continuous functions defined on compact sets, the proof is similar and continuity guarantees you the condition which is your hypothesis here (sometimes it is easier to verify that the function is continuous).

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