Completeness of the bounded-real-function-space B(S)

By | August 9, 2010

Let \(S\) be a nonempty set and \(B(S)\) a metric space of all the bounded real functions on \(S\) with the supremum norm \[\Vert f \Vert = \sup \left\{ | f(s) | ~|~ s\in S \right\} \] and the metric function \[ d(f,~g) = \Vert f-g \Vert . \] Then \(B(S)\) becomes a complete metric space.

To prove this fact, let \(\left\{ f_n \right\}\) be a Cauchy sequence in \(B(S).\) We must show that \(\left\{ f_n \right\}\) converges to a function \(f\) in \(B(S).\) This is done in two steps. Firstly, we identify a likely candidate for \(f,\) and then secondly we show that \(f\) is indeed the right choice. To identify the candidate take \(\epsilon > 0.\) Since \(\left\{ f_n \right\}\) is a Cauchy sequence, we can find \(N\) such that whenever both \(p\) and \(q\) are greater than or equal to \(N,\) then \(\Vert f_p - f_q \Vert \le \frac{\epsilon}{2} ,\) or equivalently \( | f_p (s) - f_q (s) | \le \frac{\epsilon}{2} \) for all \(s \in S.\) This means that for each fixed value of \(s\) in \(S,\) \(\left\{ f_n (s) \right\}\) is a Cauchy sequence of real numbers and so has a limit which is a real number. We will denote this number by \(f(s).\) From the triangle inequality, \[ | \Vert f_p \Vert - \Vert f_q \Vert | \le \Vert f_p - f_q \Vert , \] and so \(\left\{ \Vert f_n \Vert \right\}\) is a Cauchy sequence of real numbers. It follows that this same sequence is bounded, and from this we may deduce that \(f\) lies in \(B(S).\) Hence \(\left\{ f_n \right\}\) converges pointwise to \(f.\) We show that in fact the convergence is uniform. Fix \(p \ge N\) and \(s \in S.\) Choose \(q \ge N\) so that \( | f_p (s) - f_q (s)| \le \frac{\epsilon}{2} .\) Since \(f_n (s) \to f(s)\) as \(n \to \infty\) we may further constrain \(q\) to be sufficiently large to ensure that \( | f_q (s) - f(s)| < \frac{\epsilon}{2} .\) Then \[ | f_p (s) - f(s)| = |f_p (s) - f_q (s)| + | f_q (s) - f(s)| \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = ε. \] This establishes that for all \(s \in S\) and \( p \ge N,\) \( | f_p (s) - f(s) | \le \epsilon ,\) i.e. for all \( p \ge N,\) \(\Vert f-f_p \Vert \le \epsilon .\) Thus \(\left\{ f_n \right\}\) converges to \(f\) in \((B(S),~ \Vert \cdot \Vert ). \)

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