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.


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.

  • 1
    $\begingroup$ What do you have? $\endgroup$ – Demosthene May 16 '15 at 14:04
  • $\begingroup$ @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. $\endgroup$ – 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 $d\in [a,b]$.

  • $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Apr 6 '17 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.