Show ##\{nx^{n}(1-x)\}_{n = 0}^{\infty}## converges uni

[Euklidian-Space]

@Euklidian-Space as you asked for more detail :

To expand this last step, let's assume uniform convergence, and take $\epsilon=\frac{1}{2e}$ .
From the convergence of $f_n(x_n) \rightarrow \frac{1}{e}$ we know that
(1) $\exists n_0,\forall n\geq n_0, f_n(x_n)>\frac{1}{2e}$
But uniform convergence $f_n\rightarrow_u 0$ tells us that
(2) $\exists N, \forall n\geq N, \forall x \in [0,1], f(x)<\frac{1}{2e}$
Now take any $n\geq\max(n_0,N)$ and set $x=x_n$ .
By (1) we have $f_n(x)>\frac{1}{2e}$ , but (2) yields $f_n(x)<\frac{1}{2e}$ so we have a contradiction.

Wabbit,
i am getting $f_{n}(x_{n}) = n\left(1-\frac{n}{n+1}\right)^{n+1}$

does that still converge to $\frac{1}{e}?$

 

[wabbit]

Wabbit,
i am getting $f_{n}(x_{n}) = n\left(1-\frac{n}{n+1}\right)^{n+1}$

does that still converge to $\frac{1}{e}?$

No it doesn't, but what you're getting is incorrect - check your calculations again, you must have let a mistake slip by somewhere.

 

[WWGD]

Actually, your sequence $f_n(x)=nx^n(1-x)$ does not converge uniformly on $[0,1]$.
It converges pointwise to zero on that interval. However, if you set $x_n=\frac{n}{n+1}$ which is the value at which $f_n$ reaches its maximum over $[0,1]$ , you find that $f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0$ and this proves that the convergence is not uniform.
<Snip>.

I don't understand, aren't you saying here that the series converges pointwise to $0$ on $[0,1]$ but $f_n(x_n) \rightarrow 1/e$. How is this possible?

 

[wabbit]

I don't understand, aren't you saying here that the series converges pointwise to $0$ on $[0,1]$ but $f_n(x_n) \rightarrow 1/e$. How is this possible?

To state it in more general terms, if $f_n\rightarrow f$ and $x_n\rightarrow x$ we might expect to conclude that $f_n(x_n)\rightarrow f(x)$ - and while it is easy to prove that this is a theorem if the convergence is uniform, the conclusion does not hold if it is only pointwise.

That sequence was built I imagine precisely to illustrate how, lacking uniform convergence or another suitable assumption, the diagonal sequence may converge to a limit which is not the pointwise limit. Another very similar one which makes this perhaps clearer is the sequence of piecewise linear functions with $f_n(0)= f_n(1-\frac{2}{n})=0 ; f_n(1-\frac{1}{n})=1 ; f_n(1)=0$ and linear interpolation between those points : for any given $x$ the sequence $f_n(x)$ is constant and equal to zero except for a finite number of terms, yet by definition $f_n(x_n)=1$ for $x_n=1-\frac{1}{n}$ .

 

[wabbit]

One more comment : I suspect that for continuous functions on a compact set, the statement "$\forall (x_n)|x_n\rightarrow x, f_n(x_n)\rightarrow f(x)$" is actually equivalent to uniform convergence "$f_n\rightarrow_u f$".