- #1

- 351

- 81

*If ##f: [a,b] \to \mathcal{R}## is continuous, and there is a ##L## such that ##f(a) \lt L \lt f(b)## (or the other way round), then there exists some ##c \in [a,b]## such that ##f(c) = L##.*Proof: Let ##S = \{ x: f(x) \lt L\}##. As ##S## is a set of real numbers and non-empty, therefore we can assume ##\sup S = c##.

CASE 1: Here is the standard argument "if ##f(c) \gt L##, then by continuity ##f(c-h) \gt L## for some small ##h##". How does continuity imply that? Is it like this:

if ##f(c) \gt L##, and for some small ##h## if ##f(c-h) \lt L##, then we have

$$

\begin{align*}

f(c) - f(c-h) \gt 0 \\

\lim_{h \to 0} (f(c) - f(c-h)) \gt 0 && \textrm{and by continuity, we have} \\

f(c) - f(c) \gt 0 \\

0 \gt 0 && \textrm{which is absurd} \\

\end{align*}

$$

Therefore, for small ##h## ##f(c-h) \gt L##.