\frac{1}{u} \int_{- u}^{u} (1 - φ (t)) d t = E [\frac{1}{u} \int_{- u}^{u} (1 - e^{i tX}) d t] .

The odd imaginary part integrates to zero, while

\int_{- u}^{u} e^{i tX} d t = \frac{2 s i n ( u X )}{X} .

Therefore the expression equals

2 E (1 - \frac{s i n ( u X )}{u X}),

with the quotient interpreted continuously at

X = 0

ii

Solution

For

∣ y ∣ > 2

, one has

2 (1 - sin y / y) \geq 1

. Applying this with

y = u X

and using part (i),

P (∣ X ∣ > 2/ u) \leq 2 E (1 - \frac{sin ( u X )}{u X}) = \frac{1}{u} \int_{- u}^{u} (1 - φ (t)) d t .

Continuity of

φ

at zero and

φ (0) = 1

allow

u > 0

so small that

\frac{1}{u} \int_{- u}^{u} (1 - φ (t)) d t < ε /2.

Pointwise convergence and dominated convergence make the corresponding expression for

φ_{n}

less than

ε

for all sufficiently large

n

. Part (ii) then bounds

P (∣ X_{n} ∣ > 2/ u) \leq ε .

There are only finitely many remaining

n

, so enlarge

M \geq 2/ u

until each of their tails is also at most

ε

. This proves uniform tightness.

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