# Proving the Extreme Value Theorem

Lately I have been trying to prove the extreme value theorem using the concept of Cauchy sequences but I can't figure out how to start or where to go after I have started, and I was if anyone could lead me to proving it.

Theorem:

In calculus, the extreme value theorem states that if a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain a maximum and a minimum, each at least once.

• What do you have? – Demosthene May 16 '15 at 14:04
• @Demosthene Pretty much nothing, I've been wondering how to start and I amn't certain. However, I know how to prove that a function attains a maximum value, math.duke.edu/~cbray/Stanford/2000-2001/math41/EVTProof.pdf, but do not know know whether the same method can be extended to prove a function attains a minimum value or whether I even need to do that. – Reinhild Van Rosenú Dec 19 '15 at 0:01

f is continuous and bounded on $[a,b]$ such that $c,d\in[a,b]\subset \mathbb{R}$.

Let $\alpha :=inf\left\{f(x):x\in [a,b]\right\}$. This exists because f is continuous on the bounded interval $[a,b]$. (What is it's infimum?)

$\alpha$ denoted as the infimum of the function $f$ on $[a,b]\Rightarrow \alpha+\frac 1n$ isn't the greatest lower bound of $f$. Thus $\exists$ some mapping $x_n \rightarrow f(x_n) \ s.t. f(x_n)$ lies in between its infimum $\alpha$ and $\alpha+\frac 1n$.

Mathematically this can be notated as the inequality $\alpha \leq f(x_n) \leq \alpha+\frac 1n$.

$f$ is bounded on $[a,b]\Rightarrow$ its equivalent sequence, let's call it $a_n =\left\{x_n\right\}_{n=1}^{n= \infty } \ \forall n\in\mathbb{N}$, is also bounded on $[a,b]$. Thus by Bolzano-Weierstrass theorem, $a_n$ contains at least one convergent subsequence in the interval $[a,b]$, let's call it $a_{n_i}$ for $i\in$ of $n$, such that $\lim_{i\to\infty}{ \ (a_{n_i})}=c$

Because f is continuous, the limit of the function, or in this case, sequence, is the function of the limit. So the following results:

$$\lim_{i\to\infty}{ \ (a_{n_i})}=c\Rightarrow f(\lim_{i\to\infty}{ \ (x_{n_i}))}=f(c) \ for \ c\in [a,b]$$

Going back to our definition for $\alpha$, and applying the squeeze theorem for functions, we note:

$$\alpha \leq f(x_n) \leq \alpha+\frac 1n \Rightarrow \lim_{n\to\infty}{ (\alpha)} = \lim_{n\to\infty}{f(x_n)}=\lim_{n\to\infty}{(\alpha+\frac 1n})=\alpha$$

Noting that if a sequence or function is convergent, that all of its subsequences/subfunctions are convergent to that same limit value, we thus finally have: $$\lim_{i\to\infty}{f(x_{n_i})=\lim_{n\to\infty}f(x_n)=\alpha}$$

This implies $\alpha :=inf\left\{f(x):x\in [a,b]\right\}=f(c)$

Thus the function f achieves a minimum value $f(c)$ for $c\in [a,b]$.

To prove the maximum version, consider some number $\beta \ s.t. \ \beta :=sup\left\{f(x):x\in [a,b]\right\}=f(d)$ for $d\in [a,b]$.

• A system flag has been raised. You have posted the exact same answer twice. The system doesn't like it. Is there a particular reason for you to do this? – Jyrki Lahtonen Apr 6 '17 at 3:57