# Thread: Pointwise and uniform convergence

1. ## Pointwise and uniform convergence

From my exam this morning:
Show that $\displaystyle f_n(x) = nx(1-x)^n$ converges to f == 0 pointwise on [0, 1], but not uniformly.

There was some hint about $\displaystyle \frac{n}{n + 1}^n$ going to 1/e or something like that.

I tried to let E > 0, using the definitions and whatnot, but it didn't seem to go anywhere! It was MUCH easier on the last test, when f_n was cos(x/n)...

Anywho... if you've got a solution, great. Otherwise, I'll post the answer in September after I've seen the prof again

2. Pointwise convergence. As n increases, what happens? More and more points are included as not very far from f(x) = 0. In fact, there isn't a point that won't be included eventually.

Uniform Convergence. The problem is over on the left, near x = 0. A little elementary calculus shows the first derivative having a zero at x = 1/(1+n). This certainly moves toward x = 0. What is the value of the function at x = 1/(1+n)? Whatever it is, it's limit needs to be zero if we are to establish Uniform Convergence

It is $\displaystyle \left(\frac{n}{n+1}\right)^{n+1}$.

Use your favorite logarithm technique to prove to what this converges.

Rewriting as $\displaystyle \frac{ln\left(\frac{n}{n+1}\right)}{\frac{1}{n+1}}$ may not be obvious, but this invites l'Hospital.

A single application (and a little algebra) produces -(1 + (1/n)). Well, that's clear enough. So, the limit of the logarithm is -1. Where does that leave us?

3. Thanks for the reply. The uniform convergence argument is surely what we were expected to produce. There WAS a hint about finding f'...

I'm still not convinced about pointwise here. Do the function values approach 0 because of the n in the exponent?

4. Define $\displaystyle g(y)=yx(1-x)^y=\frac{yx}{e^{-y\ln(1-x)}}$ and take $\displaystyle y\rightarrow \infty$ (regard x as a constant).

5. Originally Posted by Jose27
Define $\displaystyle g(y)=yx(1-x)^y=\frac{yx}{e^{-y\ln(1-x)}}$ and take $\displaystyle y\rightarrow \infty$ (regard x as a constant).
Ok, that makes sense. Thanks.

I just woke up from a dream in which I got 100% on this exam...
THAT didn't happen!