Let $(f_n)_n$ be a sequence of functions defined by $$f_n : [0,1] \rightarrow \mathbb{R}: x \mapsto nx^n(1-x). $$

We have $\lim_{n \to \infty} f_n(x) = 0$ for all $x$. My textbook says the convergence isn't uniformly though, and I don't understand why.

I computed $$\sup_{x \in [0,1]} |f_n(x) - 0 | = \sup_{x \in [0,1]} | nx^n (1-x) | = \ ? $$ How do I figure out this supremum?

When I use the definition, I see that $$| nx^n(1-x) - 0 | = | nx^n - nx^{n+1}| \leq | nx^n | + | nx^{n+1}| \leq n + n = 2n. $$ And I can never get this smaller than $\epsilon$. But I couldn't find an explicit $\epsilon > 0$ and a $x \in [0,1]$ such that $ |f_n(x)| \geq \epsilon. $

Any help is appreciated.




$$f_n(x)=nx^n(1-x)\implies f'_n(x)= n^2x^{n-1}(1-x)-nx^n=nx^{n-1}\left(n(1-x)-x\right)=$$


so if $\;x\in (0,1)\;$ , we get that

$$f_n'(x)=0\iff x=\frac n{1+n}$$

and it's easy to check the above is a maximum , so

$$\forall\;x\in (0,1)\;,\;\;f_n(x)\le f_n\left(\frac n{1+n}\right)=n\left(\frac n{1+n}\right)^n\left(1-\frac n{1+n}\right)=$$

$$=\frac n{n+1}\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}e^{-1}$$

  • 1
    $\begingroup$ You just proved that $f_n(x)$ is less than an expression that tends to $e^{-1}$. But that doesn't prove that it doesn't converge uniformly (the zero function sequence has the same property). // The proof is contained in these steps, but there are several unnecessary things like checking that $f_n$ takes a maximum at the prescribed point. It is sufficient to find a sequence $x_n$ such that $f_n(x_n)\nrightarrow 0$. $\endgroup$ – Mario Carneiro Aug 25 '16 at 5:10
  • $\begingroup$ @MarioCarneiro You understand what a hint is, right? $\endgroup$ – DonAntonio Aug 25 '16 at 10:48

This is not a solution but it might be useful to others. I decided to investigate the functions geometrically, so I thought I would share in case it brings to light the methods which have been put forward by others.

enter image description here

This demonstrates (non-rigorously) that the function ought not to converge uniformly and it shows the maximum at $\frac{n}{n+1}$ approaching $\frac{1}{e}$ as $n$ grows.


Hint: Show that the extremum of $f_n$ is attained at $x_n=\frac{n}{n+1}$ and that $f_n(x_n)= n\ \left( \frac{n}{n+1} \right)^n (1-\frac{n}{n+1})$ converges as $n\rightarrow \infty$ to a non-zero value.

  • $\begingroup$ Why do I have to work with derivatives? Is there some theorem related to uniform convergence? Sorry, I only started to learn real analysis couple months ago. I only know the definition and also the equivalence with $\sup | f_n(x) - f(x)| \rightarrow 0$ as $n \to \infty$. $\endgroup$ – Kamil Aug 24 '16 at 20:38
  • $\begingroup$ You don't really have to work with derivative. It is just to show that the max of the function $f_n$ is actually computable (I gather you know how to compute its derivative?) and occurs for the cited $x_n$. After that you still need to know that $(\frac{n}{n+1})^n$ converges to a non-zero value (the remaining factors are just 'decorations'). Do you know how to do this? $\endgroup$ – H. H. Rugh Aug 24 '16 at 20:52
  • $\begingroup$ Yes, thank you. I just didn't know about this trick, that is: finding the extremum to determine the supremum. I always found these suprema by just looking at the expression. But it is nice to know. $\endgroup$ – Kamil Aug 24 '16 at 20:54

Hint. Observe that $$ f_n\left(\frac{n}{n+1} \right)=\left(1+\frac1n \right)^{-n}\cdot \frac{n}{n+1} $$ then let $n \to \infty$.

  • $\begingroup$ I like this answer because it shows that derivatives are unnecessary for the proof of non-uniform convergence itself (although they are helpful for coming up with the original choice $x=\frac n{n+1}$). $\endgroup$ – Mario Carneiro Aug 25 '16 at 5:08

For this type of problems use that the maximum is achieved either on the boundary or where the derivative is zero (this is due to compactness). Computing the derivative gives $$f_n'(x)=n^2x^{n-1}(1-x)-nx^n=(n(1-x)-x)nx^{n-1}=(n-(n+1)x)nx^{n-1}.$$ so the maximum is attained at $x_n=\frac{n}{n+1}\in(0,1)$.(The boundary points are not relevant here, as $f$ there is identically zero).

Now we have $$\sup_{x\in[0,1]}|f_n(x)|=|f_n(x_n)|,$$ but this last expression can be easily computed explicitly.

Of course the same expression holds with "$\lim_{n\to \infty}$" in front of both sides, so everything reduces to computing the limit of the right hand side: if it is zero, then the convergence will be uniform, else it won't.

  • $\begingroup$ Thank you for reply. But I still don't understand why I have to determine the extremum. Is there some theorem related to uniform convergence and achieving extrema? $\endgroup$ – Kamil Aug 24 '16 at 20:43
  • $\begingroup$ It is not a theorem: it comes from the definition, where the expression $\sup |f_n|$ comes into play. $\endgroup$ – b00n heT Aug 24 '16 at 20:44
  • $\begingroup$ So to find the supremum in these kind of problems, it is always possible to find them by computing the derivative, and checking where the function achieves its maximum? $\endgroup$ – Kamil Aug 24 '16 at 20:46
  • $\begingroup$ mostly (from my experience at least) yes (If the sequence is made of sufficiently smooth functions, which is almost always the case). Of course never forget the boundary points when looking for maxima $\endgroup$ – b00n heT Aug 24 '16 at 20:47

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