# Uniform convergence of a sequence of functions on $[0, 1]$

If $f_n(x) = n^{-x}x^n\cos(nx)$ defines the sequence of functions $\{f_n\}_{n \in \mathbb{N}}$. Does the sequence of function converge uniformly on $[0,1]$?

Should I just do the pointwise convergence, and then determine the uniform convergence by evaluating $M_n:= \sup \{ |f_n(x) - f(x)| \}$?

If taking the path of pointwise convergence, I get stuck with $f_n(1)$ as it leaves me with $\cos(n)$ on the RHS.

Thank you.

My work

$f_n(0) = 0$, $f_n(1) = 0$, and $\lim_{n \to \infty} f_n(x) = 0$ (using L'Hospital's Rule).

Then, $M_n := \sup\{|f_n(x) - f(x)|: x \in [0,1]\} = \max \{\frac{x^n\cos(nx)}{n^x}; 0 \le x \le 1\}$

$f_n'(x) = 0\Rightarrow x = 0$; so $M_n = f_n(0) = 0 \to 0$ as $n \to \infty$, obviously.

Therefore as $M_n \to 0$, as $n \to \infty$ we can conclude that $f_n$ is uniformly convergent on $[0, 1]$.

## 1 Answer

Actually the problem is not $\cos(nx)$ which is uniformly bounded by $1$, but the behaviour of the product $x^n\cdot n^{-x}$ near $0^+$ and $1^-$.

Hint. Note that if $x\in[0,1/2]$ then $$|f_n(x)|=\frac{x^n|\cos(nx)|}{n^x}\leq x^n\leq \frac{1}{2^n}.$$ On the other hand if $x\in[1/2,1]$ then $$|f_n(x)|=\frac{x^n|\cos(nx)|}{n^x}\leq \frac{1}{n^x}\leq \frac{1}{\sqrt{n}}.$$

• but what would be the justification for that split of [0, 1/2] and [1/2, 1]? Is it because we have component of cos in our function? – ISuckAtMathPleaseHELPME Dec 6 '17 at 10:24
• You have a problem at $f_n(1)$?, Then separate difficulties by considering $[0,1]$ as the union of $[0,r]$ and $[r,1]$ with $r$ any number in $(0,1)$. – Robert Z Dec 6 '17 at 10:27
• Actually the problem is not $\cos(nx)$ which is uniformly bounded. The problem is the product $x^n$ and $n^{-x}$ and its behaviour near $0^+$ and $1^-$. – Robert Z Dec 6 '17 at 10:31
• But when we do the pointwise of $f_n$ ,then at x = 1 $f_n(1) = (1)^nn^{-1}cos(1n) = (1)(1/n)(cos(n)$; unless I'm missing something obvious ... – ISuckAtMathPleaseHELPME Dec 6 '17 at 10:34
• Yes $f_n(1)=\cos(n)/n$ and by taking the limit as $n\to+\infty$, it goes to $0$. So the pointwise limit of $f_n(x)$ at $x=1$ is $0$. – Robert Z Dec 6 '17 at 10:38