Let S be a non empty, closed and bounded set (also called compact set) in $mathbb{R}^n$ and let $f:Srightarrow mathbb{R} $ be a continuous function on S, then the problem min $left { fleft ( x right ):x in S right }$ attains its minimum.

Proof

Since S is non-empty and bounded, there exists a lower bound.

$alpha =Infleft { fleft ( x right ):x in S right }$

Now let $S_j=left { x in S:alpha leq fleft ( x right ) leq alpha +delta ^jright } forall j=1,2,…$ and $delta in left ( 0,1 right )$

By the definition of infimium, $S_j$ is non-empty, for each $j$.

Choose some $x_j in S_j$ to get a sequence $left { x_j right }$ for $j=1,2,…$

Since S is bounded, the sequence is also bounded and there is a convergent subsequence $left { y_j right }$, which converges to $hat{x}$. Hence $hat{x}$ is a limit point and S is closed, therefore, $hat{x} in S$. Since f is continuous, $fleft ( y_i right )rightarrow fleft ( hat{x} right )$.

Since $alpha leq fleft ( y_i right )leq alpha+delta^k, alpha=displaystylelim_{krightarrow infty}fleft ( y_i right )=fleft ( hat{x} right )$

Thus, $hat{x}$ is the minimizing solution.

Remarks

There are two important necessary conditions for Weierstrass Theorem to hold. These are as follows −

• Step 1 − The set S should be a bounded set.

  Consider the function fleft ( x right )=x$.

  It is an unbounded set and it does have a minima at any point in its domain.

  Thus, for minima to obtain, S should be bounded.

• Step 2 − The set S should be closed.

  Consider the function $fleft ( x right )=frac{1}{x}$ in the domain left ( 0,1 right ).

  This function is not closed in the given domain and its minima also does not exist.

  Hence, for minima to obtain, S should be closed.