The answer is No, but I found it quite tricky to construct a counterexample.

Start with something easier. Define a sequence of functions

on the unit interval as follows: on the interval from 0 to

,

increases linearly from 0 to 1; on the interval from

to

,

decreases linearly from 1 to 0; and on the remainder of the interval,

. Then it is well known that

pointwise but not uniformly on the unit interval. In fact, if the binary expansion of x has a 1 in any of the first k places, then

whenever

.

Now extend

by periodicity to a function on the whole real line, with period 1. Thus

is a function with a narrow spike, of height 1, to the right of each integer point.

Next, define a sequence of functions

by

. Then

is a continuous function. If

then

. But

attains the value

in every interval of length

. Therefore

for all

in any interval

of length

. Therefore

does not tend uniformly to 0 in any interval.

However,

pointwise. To see that, we have to consider two separate cases.

__Case 1:__ x has only finitely many 1's in its binary expansion. Suppose that the last 1 occurs in the r'th place in the binary expansion, and let

. Then for

,

has a 1 somewhere in the first n places of its binary expansion, and so

. And for

,

is an integer, so again

. Thus

for all

.

__ Case 2:__ x has infinitely many 1's in its binary expansion. Then given

, there exists

with

such that x has a 1 in its

'th binary place. If

then

. But the first of those sums is 0, and the second is less than

. Therefore

.

So in either of the two cases,

as

.