Answer to Question #310913 in Real Analysis for Nikhil Singh

Question

Let fn(x)= n/(x+2n) is uniformly convergent in [0,k] , k>0

Answer 1

ANSWER Let $x\in[0,k]$ , then $\lim_{x\rightarrow\infty}\frac{n}{x+2n}=\lim_{x\rightarrow\infty}\frac{1}{\frac{x}{n}+2 }=\frac{1}{2}$ .

So, the sequence converges pointwise ( $f_{n}\rightarrow f$ ) on $[0,k]$ ,where $f(x)=\frac{1}{2}$ .

$\left | f_{n}(x)- f(x) \right |=\left |\frac{n}{x+2n}-\frac{1}{2} \right | =\left | \frac{2n-x-2n}{x+2n} \right | =\frac{ x}{x+2n}$ .

$\frac{x}{x+2n}\leq\frac{k}{2n}$ for all $x\in[0,k]$ because $x+2n\geq2n$ . Therefore, $0\leq\left | f_{n}(x)- f(x) \right | \leq\frac{k}{2n}$ and

$0\leq\underset{x\in[0,k] }{sup}\left | f_{n}(x)- f(x) \right | \leq\frac{k}{2n}.$

$\lim_{n\rightarrow\infty}\frac{k}{2n }=0$ . Hence

$\lim_{ n\rightarrow\infty}\underset{x\in[0,k] }{sup}\left | f_{n}(x)- f(x) \right | =0$

Thus, by the definition, the sequence converges uniformly on $[0,k].$