The Real And Complex Number Systems

(a) Provet that the formula df, g f g defines a metric d on BS.
Proof: We prove that d isametriconBS as follows.
(1) If df, g 0, i.e., f g sup xS |fx gx| 0 |fx gx| for all x S.
So, we have f g on S.
(2) If f g on S, then |fx gx| 0 for all x S. Thatis,f g 0 df, g.
(3) Given f, g BS, then
df, g f g
sup|fx gx|
xS
sup|gx fx|
xS
g f
dg, f.
(4) Given f, g, h BS, thensince
|fx gx| |fx hx| |hx gx|,
we have
|fx gx|
sup
xS
|fx hx| sup|hx gx|
f h h g
which implies that
f g sup|fx gx| f h h g.
xS
So,wehaveprovethatd isametriconBS.
(b) Prove that the metric space BS, d is complete. Hint: If f n is a Cauchy
sequence in BS, show that f n x is a Cauchy sequence of real numbers for each x in S.
Proof: Let f n be a Cauchy sequence on BS, d, That is, given 0, there is a
positive integer N such that as m, n N, wehave
df, g f n f m sup|f n x f m x| . *
xS
So, for every point x S, the sequence f n x R is a Cauchy sequence. Hence, the
sequence f n x is a convergent sequence, say its limit fx. It is clear that the function
fx is well-defined. Let 1 in (*), then there is a positive integer N such that as
m, n N, wehave
|f n x f m x| 1, for all x S. **
Let m , andn N, wehaveby(**)
|f N x fx| 1, for all x S
which implies that
|fx| 1 |f N x|, for all x S.
Since |f N x| BS, say its bound M, and thus we have
|fx| 1 M, for all x S
which implies that fx is bounded. That is, fx BS, d. Hence, we have proved that
BS, d is a complete metric space.
Remark: 1. We do not require that S is bounded.
2. The boundedness of a function f cannot be remove since sup norm of f is finite.
xS