- #1

Eclair_de_XII

- 1,083

- 91

- TL;DR Summary
- Let ##f## be a continuous function defined on a sequentially compact set ##K##. Then there is a point ##x_0\in K## s.t. ##f(x_0)=\sup f(K)##.

If ##f## is a constant function, then choose any point ##x_0##. For any ##x\in K##, ##f(x_0)\geq f(x)## and there is a point ##x_0\in K## s.t. ##f(x_0)=\sup f(K)=\sup\{f(x_0)\}=f(x_0)##.

Now assume that ##f## is not a constant function.

Construct a sequence of points ##x_n\in K## as follows:

1. Pick an arbitrary point ##x_1\in K##.

2. Pick a point ##x_2\in K## s.t. ##f(x_2)>f(x_1)##.

3. Repeat ad-nauseum. If there comes a point where this is impossible, then the last term of the sequence ##\{x_n\}## must be the point at which ##f## achieves its maximum.

It follows from sequential compactness that there is a subsequence ##\{x_{n_k}\}## converging to some point ##x_0\in K##. By construction, ##\{f(x_{n_k})\}## is an increasing sequence of points.

If the sequence ##\{f(x_{n_k})\}## is unbounded, then there exists an integer ##m\in \mathbb{N}## s.t. ##f(x_m)>f(x_0)##. Set ##\epsilon=f(x_m)-f(x_0)##. Then there is an integer ##N\in\mathbb{N}## s.t.:

\begin{equation}

n_k\geq N\Longrightarrow |f(x_{n_k})-f(x_0)|<\epsilon

\end{equation}

If ##m\geq N##, choose ##n_k=m##:

\begin{equation}

f(x_m)-f(x_0)>f(x_m)-f(x_0)

\end{equation}

This is a blatant contradiction.

If ##m<N##, then for ##n_k\geq N##:

\begin{eqnarray}

f(x_m)-f(x_0)>f(x_{n_k})-f(x_0)\\

f(x_m)>f(x_{n_k})

\end{eqnarray}

But ##n_k>m##, which contradicts the fact that ##\{f(x_{n_k})\}## is an increasing sequence of points.

Hence, the sequence ##\{f(x_{n_k})\}## is bounded and so has finite cardinality. The last term of the sequence, therefore, is the point at which ##f## achieves its maximum on ##K##.

Now assume that ##f## is not a constant function.

Construct a sequence of points ##x_n\in K## as follows:

1. Pick an arbitrary point ##x_1\in K##.

2. Pick a point ##x_2\in K## s.t. ##f(x_2)>f(x_1)##.

3. Repeat ad-nauseum. If there comes a point where this is impossible, then the last term of the sequence ##\{x_n\}## must be the point at which ##f## achieves its maximum.

It follows from sequential compactness that there is a subsequence ##\{x_{n_k}\}## converging to some point ##x_0\in K##. By construction, ##\{f(x_{n_k})\}## is an increasing sequence of points.

If the sequence ##\{f(x_{n_k})\}## is unbounded, then there exists an integer ##m\in \mathbb{N}## s.t. ##f(x_m)>f(x_0)##. Set ##\epsilon=f(x_m)-f(x_0)##. Then there is an integer ##N\in\mathbb{N}## s.t.:

\begin{equation}

n_k\geq N\Longrightarrow |f(x_{n_k})-f(x_0)|<\epsilon

\end{equation}

If ##m\geq N##, choose ##n_k=m##:

\begin{equation}

f(x_m)-f(x_0)>f(x_m)-f(x_0)

\end{equation}

This is a blatant contradiction.

If ##m<N##, then for ##n_k\geq N##:

\begin{eqnarray}

f(x_m)-f(x_0)>f(x_{n_k})-f(x_0)\\

f(x_m)>f(x_{n_k})

\end{eqnarray}

But ##n_k>m##, which contradicts the fact that ##\{f(x_{n_k})\}## is an increasing sequence of points.

Hence, the sequence ##\{f(x_{n_k})\}## is bounded and so has finite cardinality. The last term of the sequence, therefore, is the point at which ##f## achieves its maximum on ##K##.

Last edited: