Theorem 8.3.3: Weierstrass Convergence Theorem |
Suppose fn are a sequence of functions defined on D such thatwhere || fn ||D is the sup-norm on D. Then the (function) series |
The proof will involve two steps:
Let's get started with the first step. Fix
x
D. Then the
series
f(x)
is a numeric series. In fact, since
|f(x)|
|| f ||D that
numeric series must converge according to the
Comparison Test
to a limit L. That limit depends on x so that we
can a function
F(x) =f(x)
which is well-defined (pointwise) for all
x
D.
Next we need to show that the convergence is uniform. Take any
> 0. Since the numeric
series
|| fn ||D
converges we can find an integer N such that
<
for all n > N
But then:
which means that the sequence of partial sums converges uniformly.
If we knew what uniformly Cauchy meant, we could also show that the sequence of partial sums was uniformly Cauchy, which (we would hope) should imply uniform convergence. To practice, define uniformly Cauchy sequences and finish the proof that way.
If, in addition, each fn is continuous, then every N-partial sum is also continuous. Since uniform convergence preserves continuity, the limit would also be continuous.
