Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 5 / a

S_{N} (x) = \sum_{n = 1}^{N} f_{n} (x)

s : E \to R

\forall ε > 0 \exists N_{0} \in N \forall N \geq N_{0} \forall x \in E : ∣ S_{N} (x) - s (x) ∣ < ε .

The same threshold

N_{0}

must control every point of

E

. Equivalently, the supremum error

sup_{x \in E} ∣ S_{N} (x) - s (x) ∣

tends to zero, for nonempty

E

ii

f_{n} : E \to R

C

satisfy

∣ f_{n} (x) ∣ \leq M_{n} for every x \in E, M_{n} \geq 0, \sum_{n = 1}^{\infty} M_{n} < \infty,

then

\sum f_{n}

E

\sum ∣ f_{n} ∣

also has uniform convergence, since every tail is bounded by the corresponding tail of

\sum M_{n}

x \in E sup n = N + 1 \sum \infty ∣ f_{n} (x) ∣ \leq n = N + 1 \sum \infty M_{n} ⟶ 0.

Take

0 \leq r < R

. At the point

r

\sum_{n = 0}^{\infty} ∣ a_{n} ∣ r^{n} < \infty

. On the whole interval

[- r, r]

∣ a_{n} x^{n} ∣ \leq ∣ a_{n} ∣ r^{n} .

M_{n} = ∣ a_{n} ∣ r^{n}

proves uniform convergence on $[- r, r]$ . It also gives the explicit tail estimate

∣ x ∣ \leq r sup n = N + 1 \sum \infty a_{n} x^{n} \leq n = N + 1 \sum \infty ∣ a_{n} ∣ r^{n} ⟶ 0.

For

r = 0

the interval consists only of zero, and all positive-degree terms vanish. If

R = \infty

, the same argument applies to every finite

r

; if

R = 0

, there is no

r \geq 0

with

r < R

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