Alternative Proof of the Extreme Value Theorem

I have proven the Boundedness Theorem for continuous functions and would now like to prove the Extreme Value Theorem; that is, show that the upper bound is indeed attained for continuous functions. I would like to use the most direct approach as possible, straight from the properties of real continuous functions (the approach in Spivak's Calculus seems too clever for me). I also want to avoid using sequences.

Assuming that $f$ is continuous and bounded on $[a,b]$, here's what I've done so far:

Let $g(x) = \sup \{f(t), t \in [a,x] \}$ and $A = \{x : x \in [a,b] \text{ and } g(x)=g(b)\}$. Let $\alpha = \inf A$, then $\alpha \in [a,b]$, so $f(\alpha)$ is defined. Also $f(\alpha) \leq g(b)$ since $g(b) = \sup \{f(t) : t \in [a,b]\}$. So we only need to show that $f(\alpha) < g(b)$ entails a contradiction.

Assume $f(\alpha) < g(b)$, then $g(b)-f(\alpha) > 0$. Then by continuity of $f$ at $\alpha$, there is a $\delta$, such that $f(x)-f(\alpha) < g(b) - f(\alpha)$ and so $f(x) < g(b)$ for all $x \in [\alpha-\delta, \alpha + \delta]$. But since $\alpha$ is a greatest upper bound, there is an $x' \in [\alpha, \alpha + \delta]$ such that $x' \in A$ and so $g(x') = g(b)$.

Can this proof be finished or am I just headed for nowhere?

• Your proof is correct and you have already finished the proof. There are minor typos. You need to use open intervals in last paragraph like $(\alpha-\delta, \alpha + \delta)$ and $(\alpha, \alpha + \delta)$. Rest is OK. – Paramanand Singh Jun 30 '15 at 5:28
• You may also use the nested interval principle. Let $M$ be supremum of $f$ on $[a, b]$. Then $M$ must be the sup on at least one of $[a, c]$ and $[c, b]$ where $c = (a + b)/2$. Call that interval as $[a_{1}, b_{1}]$. Repeat the same and you get a nested sequence of intervals $[a_{n}, b_{n}]$ such that there is a unique $\alpha$ which lies in all these nested intervals. Prove $f(\alpha) = M$. For a survey of various proofs of properties of continuous functions see my posts starting from paramanands.blogspot.com/2011/06/… – Paramanand Singh Jun 30 '15 at 5:35
• But I still need to show that the fact that there's an $x' \in [\alpha, \alpha+\delta]$ such that $g(x')=g(b)$ contradicts the fact that $f(x)<g(b)$ for all $x \in [\alpha-\delta, \alpha+\delta]$. It doesn't seem quite trivial to me. – Simeon Jun 30 '15 at 9:07
• Thats really trivial. Since $\alpha - \delta < \alpha$ it is clear that $\alpha - \delta \notin A$. In fact all the numbers of $[a, \alpha - \delta]$ are not in $A$. and hence $g(x) < g(b)$ for all $x \in [a, \alpha - \delta]$ and since $f(x) < g(b)$ in $[\alpha - \delta, \alpha + \delta]$ it follows that $g(x) < g(b)$ for all $x \in [a, \alpha + \delta]$. Now you get the contradiction. – Paramanand Singh Jun 30 '15 at 9:18
• I see, but how does it follow that if $f(x) < g(b)$ in $[a-\delta, a+\delta]$, then $g(x) < g(b)$ in $[a-\delta, a+\delta]$? $g$ gives the upper bound of $f$, not the maximum value. – Simeon Jun 30 '15 at 9:28

Let's use the notation provided by OP. Thus we have $$g(x) = \sup\,\{f(t)\mid t \in [a, x]\},\,\, A = \inf\,\{x\mid x \in [a, b], g(x) = g(b)\},\,\, \alpha = \inf\, A$$ Clearly we can see that $M = g(b)$ is the supremum of $f$ on whole interval $[a, b]$ and $\alpha \in [a, b]$. And our objective is to find a point $x \in [a, b]$ such that $f(x) = M$.
If $f(a) = M$ then we are done. Hence let's assume that $f(a) < M$. This ensures that there is a range of values of $x$ near $a$ where $f(x) < M$ and hence we have an $h > 0$ such that $f(x) < (M + f(a))/2$ for all $x \in [a, a + h]$ (this is derived by using continuity of $f$ at $a$ using the value of $\epsilon = (M - f(a))/2$). This implies that $g(x) < M$ for all $x \in [a, a + h]$. So $A$ does not contain any point of $[a, a + h]$ and hence $\alpha > a$. Same way we can assume that $f(b) \neq M$ and deduce that $\alpha < b$. Thus $\alpha \in (a, b)$.
We now claim that $f(\alpha) = M = g(b)$. Suppose that it is not the case then $f(\alpha) < M$ and let's put $\epsilon = (M - f(\alpha))/2$. Then we have a $\delta > 0$ such that $(\alpha - \delta, \alpha + \delta) \subseteq [a, b]$ and $$f(\alpha) - \epsilon < f(x) < f(\alpha) + \epsilon$$ for all $x \in (\alpha - \delta, \alpha + \delta)$. It follows that $$K = \sup\,\{f(x) \mid x \in (\alpha - \delta, \alpha + \delta)\}\leq f(\alpha) + \epsilon = \frac{M + f(\alpha)}{2} < M\tag{1}$$ Consider the number $\beta = \alpha -(\delta/2)$. Clearly $\alpha - \delta < \beta < \alpha$ and hence $\beta \notin A$. This means that $g(\beta) < g(b) = M$. Thus supremum of $f$ on interval $[a, \beta]$ is less than $M$ and from equation supremum of $f$ on $(\alpha - \delta, \alpha + \delta)$ is also less than $M$. Since these two intervals have an interior point in common it follows that supremum of $f$ on their union $[a, \alpha + \delta)$ is also less than $M$.
Now by definition of $\alpha = \inf\, A$ it follows that there is a $\gamma \in A$ such that $\alpha < \gamma < \alpha + \delta$ and hence supremum of $f$ on $[a, \gamma]$ is $g(b) = M$. This is contrary to the fact established in last paragraph (because $\gamma$ is interior point of $[a, a + \delta)$).
The contradiction shows that $f(\alpha) = M = g(b)$.
• Aha! I just needed to use $\frac{M-f(a)}{2}$ as my epsilon. – Simeon Jun 30 '15 at 11:05
• @Simeon: Now you understand the trick... Actually we need to limit the values of $f$ to something less than a number $K$ and this $K$ needs to be any number satisfying $f(a) < K < M$. This can be done by choosing $0 < \epsilon < M - f(a)$. We have chosen a specific $\epsilon$ which ensures that $K$ lies halfway between $M$ and $f(a)$. But you can choose any value of $\epsilon$ with $0 < \epsilon < M - f(a)$. – Paramanand Singh Jun 30 '15 at 11:09