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Question: If $f$ is increasing on $(a,b)$, continuous at $a$, $b$, then $f$ is increasing on $[a,b]$.

My Work:

It remains to prove that $f(a) < f(x) < f(b)$ for $x \in (a,b)$.

Assume the contrary and without loss of generality, assume there exists $x_0 \in (a,b)$ such that $f(a) \geq f(x_0)$.

I have proved when $f(a) > f(x) \implies f(a) - f(x) > 0$

Let $\varepsilon = f(a) - f(x)$ and $\displaystyle x = \min\left(\frac{\delta}{2}+ a, \frac{x + x_0}{2}\right) \implies f(x_0) > f(x)$ since $f$ is increasing on $(a,b)$.

Then $\forall \delta > 0$

If $0 < x - a < \delta$ then $\varepsilon = f(a) - f(x_0) < f(a) - f(x) \implies |f(x) - f(a)| > \varepsilon$.

Therefore, $f$ is not continuous at $a$. Contradiction.

Is there anything wrong with this proof?

Also, how would you get the contradiction if $f(a) = f(x_0)$ for some $x_0 \in (a,b)$?

Setting $\varepsilon = f(a) - f(x_0) + 1$ does not do much.

Thanks.

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2 Answers 2

Note that we have two definition for increasing or decreasing. for example $f$ is increasing in $[a,b]$ if for every $a\leq x_{1}\leq x_{2}\leq b$ we have $f(x_{1})\leq f(x_{2})$, another definition is strictly increasing (or strictly decreasing by same way) $f$ is strictly increasing if for every $a\leq x_{1}<x_{2}\leq b$ we have $f(x_{1})<f(x_{2})$ .

For increasing: Assume that $f(a)=f(x_{0})$. If $f$ be constant in interval $[a,x_{0}]$ then it is increasing at $a$ . but if not as $f$ is increasing in $(a,x_{0}]$ there is a point between them that the value of $f$ there is less than $f(a)$ and by previous part claim is proved.

For strictly increasing: Assume that $f(a)=f(x_{0})$. If $f$ be constant in interval $[a,x_{0}]$ then we have $f(\frac{a+x_{0}}{2})=f(x_{0})$ but $a<\frac{a+x_{0}}{2}<x_{0}<b$ that is contradiction with $f$ is strictly increasing on $(a,b)$ so $f$ is not constant between $a$ and $x_{0}$, as $f$ is strictly increasing in $(a,x_{0}]$ there is a point between them that the value of $f$ there is less than $f(a)$ and by previous part claim is proved. Why? because name that new point $x_{1}$ so $f(x_{1})<f(x_{0})=f(a)$, do your work on $a$ and $x_{1}$ instead of $x_{0}$ at this level.

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I recently asked whether my proof for this same question was acceptable here. It seems that the following is a good way in which to go about this:

Suppose there is a point $\gamma$ such that $f(a) > f(\gamma)$. By continuity of $f$ at $a$, we know that there exists some $\delta > 0$ such that $f(x) > f(\gamma)$ for all $x$ in $(a, a + \delta)$. For some such $x$ we have $x < \gamma$ and $f(x) > f(\gamma)$. This is a contradiction, so we must have $f(a) \leq f(\gamma)$ for all $\gamma \in (a, b)$.

Now suppose that there was some $\mu \in (a, b)$ such that $f(\mu) = f(a)$. Then there exists some $\mu'$ such that $a < \mu' < \mu$. Because $\mu' < \mu$ we must have $f(\mu') < f(\mu)$. But then $f(\mu') < f(a)$. This contradicts the previous paragraph. Hence $f(a) < f(\mu)$ for all $\mu \in (a, b)$.

The argument for $f(b) > f(x)$ for all $x \in (a, b)$ follows by a similar line of reasoning.

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