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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!

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    $\begingroup$ $M_n$ is increasing. $\endgroup$
    – RRL
    Jun 15 '20 at 12:54
  • $\begingroup$ 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, $\endgroup$
    – coffeemath
    Jun 15 '20 at 12:54
  • $\begingroup$ 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. $\endgroup$
    – mathdoge
    Jun 15 '20 at 12:56
  • $\begingroup$ @RRL I am actually also suspecting it is increasing, let me check. $\endgroup$ Jun 15 '20 at 13:01
  • $\begingroup$ @coffeemath yeah. I am suspecting so, I will check! Thank you! $\endgroup$ Jun 15 '20 at 13:01
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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, $M_n = \left(\frac{n}{n+1}\right)^{n+1} = \frac{1}{(1+1/n)^{n+1}} \nearrow e^{-1}$.

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

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.

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