# Question on uniform convergence

1. Aug 19, 2007

### Simfish

5: Let p > 0. Let $$f_n (x) = x n^p e^{-nx}$$

(i) For what values of p is$$f_n (x) -> 0$$ uniformly on [0,1]?
(ii) For what values of p is it true that $$lim_{n->\infty} \int_0^1 f_n (x) dx = 0$$

For (i), we know that uniform convergence implies that the convergence to 0 depends ONLY on the value of N, so it should converge to 0 for all x. Now, we can make our function arbitrarily small for any p, if p < 1,since then for n -> infinity, both functions converge to 0.

Let's try (i) first.

We want to find the maximum value of $$f_n (x)$$. So...
\begin{align} \left[ f_n (x) \right]' = \left[- x n^{p+1} e^{-nx} + n^p e^{-nx}\right] = 0 \\ \left[ e^{nx} n^{p} (-nx + 1) \right] = 0 \\ -nx + 1 = 0 \\ nx = 1 \\ x = 1/n,\\ \end{align}

plug back into equation. the value of the maximum is... $$\left[ n^{p-1} * \frac{1}{e} \right]$$
which can only imply uniform convergence for all x for p $$<$$ 1.

But what of (ii)? Integration by parts.

$$x n^p \left[\frac{-x e^{-nx}}{n} - \frac{e^{-nx}}{n^2} \right]_{0}^{1} = \left[\frac{-e^{-n}}{n} - \frac{e^{-n}}{n^2} + \frac{1}{n^2} \right] x n^p$$

...which would imply convergence of the integral for all p < 2. Am I wrong anywhere? Someone else said that p > 0, but this relation seems to be completely relevant for all negative values of p as well.

Last edited: Aug 19, 2007