Theorem. Let \(f:\left[a,b\right]\rightarrow\mathbb{R}\) be a continuous function from \(\left[a,b\right]\) to \(\mathbb{R}\).
Then \(f\) attains a global maximum and a global minimum on \(\left[a,b\right]\). \[\exists c,d\in\left[a,b\right]\ :\ \forall x\in\left[a,b\right]\quad f(c)\le f(x)\le f(d)\]
Proof. We will first prove that \(f\) attains a global maximum.
According to the Weierstrass Boundedness Theorem, \(f\) is bounded on \(\left[a,b\right]\).
Hence \(\operatorname{Im}(f)\) is a bounded subset of \(\mathbb{R}\) and it is also a non-empty set (since \(f(a)\in\operatorname{Im}(f)\)).
The upper bound theorem states that every non-empty subset of \(\mathbb{R}\) which is bounded from above has a supremum, so \(\operatorname{Im}(f)\) has a supremum.
Denote \(\beta=\sup\left(\operatorname{Im}(f)\right)\).
Also recall that since \(\beta\) is a supremum, it fulfills the supremum characterization \[\forall\varepsilon>0\ \exists y\in\operatorname{Im}(f)\ :\ \beta-\varepsilon<y\le\beta\] And in particular, \[\forall n\in\mathbb{N}\ \exists y_{n}\in\operatorname{Im}(f)\ :\ \beta-\frac{1}{n}<y_{n}\le\beta\] \[\therefore\ \forall n\in\mathbb{N}\ \exists x_{n}\in\left[a,b\right]\ :\ \beta-\frac{1}{n}<f(x_{n})\le\beta\] We shall examine \(\left(x_{n}\right)_{n=1}^{\infty}\).
It is a bounded sequence, therefore according to the Bolzano-Weierstrass theorem, it has a convergent subsequence \(\left(x_{n_{k}}\right)_{k=1}^{\infty}\).
Denote \(\underset{k\rightarrow\infty}{\lim}x_{n_{k}}=d\).
Since limits of sequences weakly preserve inequalities maintained by all elements of the sequence, \(d\in\left[a,b\right]\).
Because \(f\) is continuous, Heine’s characterization of continuous functions dictates \(\underset{k\rightarrow\infty}{\lim}f\left(x_{n_{k}}\right)=f(d)\).
Moreover, let us notice that since \(n_{k}\ge k\) we get \[\forall k\in\mathbb{N},\ \beta\ge f\left(x_{n_{k}}\right)>\beta-\frac{1}{n_{k}}\ge\beta-\frac{1}{k}\]
Since \(\underset{k\rightarrow\infty}{\lim}\left(\beta-\frac{1}{k}\right)=\beta\), the squeeze theorem implies \[f(d)=\underset{k\rightarrow\infty}{\lim}f\left(x_{n_{k}}\right)=\beta\]
Recall that a supremum which belongs to the set is always the maximum of the set, hence \(\beta=\max\left(\operatorname{Im}(f)\right)\) and \(d\) is a global maximum point in \(\left[a,b\right]\), as needed.
As for the global minimum: we know that a scalar multiple of a continuous function is also continuous, hence \(-f\) is a continuous function, and the global maximum result applies to it: \[\exists c\in\left[a,b\right]\ :\ \forall x\in\left[a,b\right]\quad-f(x)\le-f(c)\]
\[\therefore\quad\exists c\in\left[a,b\right]\ :\ \forall x\in\left[a,b\right]\quad f(x)\ge f(c)\]
This gives us the global minimum result. ∎