EDIT: I think I might have my answer now: the maximums $M_n$ are increasing with each subsequent. So the limit of $M_n$, $\large \frac{1}{e}$, is indeed the uniform bound, I believe.
The sequence is $f_n(x) = nx(1-x)^n$ for $n=1,2,3,...$ and $x \in [0,1]$.
I found that the maximizer is $x_n$ = $ \frac {1}{n+1}$, by solving for $f_n'(x)=0$, and noting that the second derivative $f_n''(x)$ is negative at the critical point $\frac {1}{n+1}$.
Then evaluating the original sequence of functions $f_n(x)$ at $x_n$ gives the maximum
$$M_n = (\frac{1}{1+\frac{1}{n}})^{n+1} $$
Now, the limit of $x_n$ = $0$ and the limit of $M_n$ = $\large \frac{1}{e}$
The part of the question that I am stuck on is:
Show that in the interval $0\le x \le 1$ the functions $f_n(x)$ are uniformly bounded. Does $$\lim_{n \to \infty} \int_0^1 f_n(x)dx = \int_0^1\lim_{n \to \infty}f_n(x)dx$$
The solution gives an answer that I'm concerned may not be correct. It says that "since $\large \frac{1}{e}$ is the maximum on [0,1], then it uniformly bounds $f_n(x)$.
Is this correct? I think that the maximum that I found above is of course a maximum over [0,1], but it depends on n. So for every n fixed, $f_n(x)$ has a maximum $M_n$ over [0,1]. I don't see why the limit of $M_n$ = $\frac {1}{e}$ would be a max for all n, and hence uniformly bounding the sequence of functions.
Also, I guess answering the above question will allow me to answer the question of the equality above, and justifying taking the limit inside of the integral. I need to use Dominated Convergence Theorem and show that $f_n(x)$ is bounded (for all n) by an integrable function. I know that I cannot use the weaker uniform convergence claim to justify moving the limit inside, since the convergence is not uniform.
Any comments or suggestions are welcome.
Thanks,
You've shown that $$\lim_{n\to\infty}\sup_{x\in[0,1]}f_n(x)=e^{-1}, $$ however $f_n$ converges pointwise to $0$. Therefore $f_n$ does not converge uniformly, as you correctly concluded. However, for each $n$ we have $\sup_{x\in[0,1]}f_n(x)= \left(1-\frac1{n+1}\right)^{n+1}$, and $\left(1-\frac1{n+1}\right)^{n+1}$ increases monotonically to $e^{-1}$. Therefore $\sup_{x\in[0,1]}f_n\leqslant e^{-1}$ and $$\int_0^1 e^{-1}\ \mathsf dx = e^{-1}<\infty, $$ so by dominated convergence it follows that $$\lim_{n\to\infty}\int_0^1 f_n(x)\ \mathsf dx = \int_0^1 \lim_{n\to\infty} f_n(x)\ \mathsf dx = 0. $$