Science & Mathematics » Mathematics » F_n(x) is a real-valued function defined on [0, 1] by the formula below. Prove that {f_n} converges pointwise.?

# F_n(x) is a real-valued function defined on [0, 1] by the formula below. Prove that {f_n} converges pointwise.?

f_n= 0 if 0<= x <2^(-n)
= 2^(n/2) if 2^(-n)<= x <= 2^(1-n)
= 0 if 2^(1-n)< x <= 1

I'm supposed to prove that {f_n} converges pointwise to 0 but the limit f_n doesn't equal 0.

I get that when x is between 2^-n and 2^(1-n), the interval is shrinking and shifting to the left as n approaches infinity. My TA gave me a hint that I'm supposed to show that given any x, I can find N such that for n>N the interval will not include x. I still need more help with solving this problem, though...

• f_n does converge pointwise, and the pointwise limit of this function is indeed f(x) = 0.

In order to show that the limit of f_n(x) is zero, consider the following two cases:

x ≠ 0:
then, as you've stated, we can find an N such that whenever n>N, 2^-n and 2^(1-n) are both less than x (more precisely, we can choose N > 1 - log(x)/log(2) for any non-zero x). Since f_n(x) = 0 for infinitely many values, we conclude that f_n(x) converges (pointwise) to 0 for any non-zero x

x = 0:
For any n, 0 will fall into the interval 0<= x <2^(-n). It is thus clear that f_n(0) approaches 0 as n->∞.

I'm not sure what you meant with your statement "the limit f_n doesn't equal 0". Do you mean you have to prove that it doesn't converge uniformly?
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