Help in proving that $\bigl (n^\alpha x^n (1-x) \bigr)_{n=0}^{\infty}$ is convergent.

Question

I need some help in the following question:

Let $\ 0<\alpha \in \mathbb R$. If $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ a sequence of functions as $\forall n\in\mathbb N \ \ \ f_n:[0,1]\to \mathbb R$:$$\forall n\in\mathbb N, \ \ \forall x\in\mathbb [0,1]: \ \ \ f_n(x)=n^\alpha x^n (1-x)$$

Q: 1. Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is pointwise convergent in $[0,1]$.

2. Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is uniformly convergent iff $\alpha<1$.

My thought till now: For the first part I tried to prove $\lim_{n\to\infty}n^\alpha x^n (1-x)=0$ for a constant $x$ but the $n^\alpha$ is confusing me and I don't know how to continue.

Thanks in advance

Answer 1

We can write estimation: $$|f_n(x)|=|n^\alpha x^n (1-x)| \leqslant n^\alpha$$ from which for $\alpha < 0$ you have as 1 so 2.

For case $0 < \alpha < 1$ It's easy find, that we have maximum in $x=\frac{n}{n+1}$ and then use criterion $$\lim_{n \to \infty} \sup_{}|n^\alpha x^n (1-x)|=\lim_{n \to \infty} n^{\alpha} \left( \frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}=0$$

Answer 2

Hint:

Just show the logarithm of $$\ln(n^\alpha x^n)=\alpha\ln n+n\ln x \to -\infty.$$

Answer 3

If $x=1$ then $f_n(x)=0$ for all $n$. Suppose $0 \leq x <1$. Let $a=-\ln x$. Then $a>0$ and $n^{\alpha} x^{n}=n^{\alpha} e^{-na}$. Hence $n^{\alpha} x^{n}(1-x)\leq e^{-na} n^{k}$ where $k$ is an integer exceeding $\alpha$. A well known consequence of L'Hopital's Rule is $e^{-na} n^{k} \to 0$ for any positive integer $k$ if $a>0$. Hence $f_n(x) \to 0$ for every $x$.

The maximum of $f_n$ is attained when $nx^{n-1}-(n+1)x^{n}=0$ or $x =\frac n {n+1}$. The maximum value is $n^{\alpha} (\frac 1 {1+\frac 1 n})^{n} \frac 1 {n+1}$. Can you show that this tends to $0$ iff $\alpha <1$?

Answer 4

A different approach for the pointwise convergence: (the standard being using a continuous variable $t\to\infty$ instead of $n$ and using DLH)

To see that $f_n$ is pointwise convergent, fix $x\in[0,1]$. If $x=1$ or $x=0$, then $f_n(x)=0$ for all $n$, so $f_n(x)$ does indeed converge (to 0). If $x\in(0,1)$, we will show that $n^\alpha x^n\to0$, then obviously $f_n(x)\to 0$ as well. Let $m$ denote the least integer with $\alpha\leq m$. Then $$\sum_{n=0}^\infty n^m x^n\sim\sum_{n=m}^\infty n(n-1)\cdots(n-m+1)x^{n-m+1}$$

which is summable, as the $m$-th derivative of the geometric series. Thus $n^mx^n\to0$, so it is also $n^\alpha x^n\to0$.

Now to check uniform convergence, observe that the supremum of $f_n$ is attained at $x_n=\frac{n}{n+1}$. Compute $f_n(x_n)$. It is $\sup_{x\in[0,1]}|f_n(x)-0|=\sup_{x\in[0,1]}|f_n(x)|=f_n(x_n)$. You will see that $f_n(x_n)\to0$ if and only if $\alpha<1$.

Answer 5

1)Pointwise convergence:

$f_n(0)=f_n(1)=0;$

$x \in (0,1)$: Root test;

$\lim_{n}\sqrt [n]{f_n(x)}=$

$\lim_{n}(1-x)^{1/n}n^{a/n}x=x <1.$

Hence

$ \lim_{n}f_n(x)=0$ for $x \in [0,1].$

2)$0<a<1;$

$x \in (0,1):$ $x=\frac{1}{1+y}$, $y >0$;

$f_n(x)=n^a\frac{1}{(1+y)^n}(1-x) \lt$

$\left (\frac{n^a}{ny}\right )\frac{y}{1+y}<\frac{1}{n^{1-a}};$

$\sup_{x}f_n(x) \le \frac{1}{n^{1-a}};$

Thus $\lim_{n}\sup_{x} =0$, uniformly convergent for 0<a<1 in $[0,1]$.

Used: $(1+y)^n >1+yn$, $y>0$, binomial expansion.