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I have a function which is \begin{equation} F(x)= \begin{cases} f(x) & x \in [\underline{x},\bar{x})\\ \\ f(\bar{x}) & x=\bar{x} \end{cases} \end{equation}

The function $f(x)$ is strictly increasing in $[\underline{x},\bar{x})$, and $\lim_{x\to \bar{x}}f(x)=f(\bar{x})$. So I could conclude that $F(x)$ is continuous at $\bar{x}$.

Could I conclude that $f(\bar{x})>f(x)$ for all $x \in [\underline{x},\bar{x})$ ?

congmingniao
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    Yes. The function is strictly increasing up to $\bar{x}$, and is continuous there (its limit is its value). Thus its strictly increasing on the closed interval. – DRF Jul 01 '15 at 12:57
  • I feel a little confused. Based on the definition of limiting. $\lim_{x \to \bar{x}}f(x)=f(\bar{x})$ is equivalent to the following statement that for any $\epsilon$, I can find a positive $\delta$ such that when $\bar{x}-x<\delta$, $|f(\bar{x})-f{x})|<\epsilon$. This statement does not say that $f(\bar{x})>f(x)$ .Then how to prove $f(\bar{x})>f(x)$ ? – congmingniao Jul 01 '15 at 13:02
  • The inequality doesn't follow from the definition of the limit in the sense you seem to want it to. It follows from the fact that if $f(\bar{x})$ is the limit of a function that is strictly increasing on an open interval terminating at $\bar{x}$ and the function is continuous at $\bar{x}$ then the continuity implies it's strictly increasing on the closed interval. This is just because if it wasn't (strictly increasing) then there would be some $c$ in the interval which is above $f(\bar{x})$ but then all the values from $f^{-1}(c)$ to $\bar{x}$ would be too large and you would lose continuity. – DRF Jul 01 '15 at 13:07
  • I see your point. Great insight ! Thank you very much. – congmingniao Jul 01 '15 at 13:11

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Yes. I will use $[a,b]$ instead of $[\underline{x},\overline{x}]$ in this demonstration.

Suppose there exist a $x \in [a,b[$ such that $f(x) \geq f(b)$

For every $\epsilon > 0$ such that $ \frac{b-x}{2} > \epsilon $, you have that

$$f(b-\epsilon) > f\left(\frac{b+x}{2}\right) > f(x) \geq f(b)$$

Now you make $\epsilon \to 0$, and you get

$$f(b) \geq f\left(\frac{b+x}{2}\right) > f(x) \geq f(b)$$

Contradiction

Tryss
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