# Show that $f_{n}(x):=nx(1-x)^{n}$ is uniformly bounded on $[0,1]$ for all $n\geq 1$.

Consider $$f_{n}(x):=nx(1-x)^{n}$$ defined for $$n=1,2,3,\cdots$$ and $$x\in [0,1]$$. The exercises have two parts:

(a) Show that for each $$n$$, $$f_{n}(x)$$ has a unique maximum $$M_{n}$$ at $$x=x_{n}$$. Compute the limit of $$M_{n}$$ and $$x_{n}$$ as $$n\rightarrow\infty.$$

(b) Prove that $$f_{n}(x)$$ is uniformly bounded in $$[0,1]$$.

I have computed that for each $$n$$, $$f_{n}(x)$$ has a unique maximum in $$[0,1]$$ at $$x_{n}=\dfrac{1}{1+n}$$ with the maximum value $$M_{n}=\Big(\dfrac{n}{n+1}\Big)^{n+1}.$$ Thus $$\lim_{n\rightarrow\infty}x_{n}=0\ \ \ \text{and}\ \ \ \lim_{n\rightarrow\infty}M_{n}=\lim_{n\rightarrow\infty}\Big(1-\dfrac{1}{1+n}\Big)^{n+1}=e^{-1}.$$

The solution says that since $$|f_{n}(x)|\leq |M_{n}|$$ for each $$n=1,2,\cdots$$, the above shows that $$|f_{n}(x)|\leq e^{-1}$$ for all $$x\in [0,1]$$ and $$n=1,2,\cdots$$.

I don't understand this. To show the uniform boundedness, don't we need to show $$\sup_{n}|f_{n}(x)|\leq C,\ \ \text{for some constant}\ C\ \text{and for all}\ x\in [0,1]?$$

It is true that since $$|f_{n}(x)|\leq M_{n}$$ for each $$n$$ and for all $$x\in [0,1]$$, we have $$\sup_{n}|f_{n}(x)|\leq\sup_{n}|M_{n}|,$$ but why does the limit of $$M_{n}$$ being $$e^{-1}$$ implies the sup is $$e^{-1}$$?

Thank you!

• $M_n$ is increasing.
– RRL
Jun 15 '20 at 12:54
• Is $(1/1-(n+1))^(n+1)$ a monotone increasing sequence as $n$ is taken $n=1,2,3,\cdots$? If so that should be easily checked, and I think would imply what you want, Jun 15 '20 at 12:54
• If you just want a uniform bound, and do not care about the precise value of the bound, then the convergence of $M_n$ already tells it: $M_n$ converges $\implies |M_n| \leq C$ for some $C > 0$. You said that the upper bound could be $e^{-1}$, I guess you could check if $M_n$ is monotone. Jun 15 '20 at 12:56
• @RRL I am actually also suspecting it is increasing, let me check. Jun 15 '20 at 13:01
• @coffeemath yeah. I am suspecting so, I will check! Thank you! Jun 15 '20 at 13:01

It is well known that $$(1 + 1/n)^n \nearrow e$$ and $$(1+1/n)^{n+1} \searrow e.$$ See one of the many proofs on this site here.

Thus, $$M_n = \left(\frac{n}{n+1}\right)^{n+1} = \frac{1}{(1+1/n)^{n+1}} \nearrow e^{-1}$$.

$$\log(M_n) = (n+1)\log\left(\frac{n}{n+1}\right)$$

And also

$$\frac{d}{dx}\left((x+1)\log\left(\frac{x}{x+1}\right)\right) = \log\left(\frac{x}{x+1}\right)+ \frac{1}{x}.$$

Since

\begin{align*}\frac{x}{x+1}e^{1/x} &= \frac{x}{x+1}\sum_{n=0}^{\infty}\frac{(\frac{1}{x})^n}{n!} \\ &= \frac{x}{x+1}(1+1/x+1/2x^2+\ldots) \\ &=\frac{x}{x+1} + \frac{1}{x+1} + \frac{1}{2x(x+1)}+\ldots \\ & = 1 + \frac{1}{2x(x+1)}+\ldots > 1, \end{align*}

so we must have that $$\log\left(\frac{x}{x+1}\right)+ \frac{1}{x} > 0$$, i.e. $$M_n$$ is increasing so $$f_n$$ is uniformly bounded by $$\lim_{n\rightarrow \infty}M_n = 1/e$$. Of course we are assuming $$x > 0$$, but that is not a problem here.