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

Let $f_n (x) =\frac{ x}{1+ nx^2}$ where $x \in [0,1]$

Then, $\lim f_n(x) = 0$ as $n \to \infty$

and so that $\langle f_n (x) \rangle$ converges pointwise to function $f(x) = 0$ on $[0,1]$.

Further by "theorem", "let $D$ subset of $R$ and $\langle f_n \rangle$ be sequence of functions defined on $D$ which converges pointwise to function $f$ on $D$ and let $M_n = \sup | f_n(x) - f(x)|$ on $D$ then $\langle f_n \rangle$ converges uniformly to function $f$ on $D$ if and only if $\lim M_n = 0$ "

By above theorem, we can see our the sequence $\langle f_n(x) \rangle$ converges uniformly to function $f(x)= 0$ on $D$. But if I go through definition of uniform convergence then I saw, for $x \in [0,1]$ $|f_n(x) - f(x)| = \frac{ x}{1+ nx^2} < 1/nx$ Hence

$|f_n(x) - f(x)| < \varepsilon$ if and only if(iff) $1/nx < \varepsilon$

iff $n > 1/x \varepsilon$

So by taking $k = [1/x \varepsilon] + 1$ we get

$|f_n(x) - f(x)| < \varepsilon$ for all $n ≥ k$

But $k$ here depends on both $x$ and $\varepsilon$. So $\langle f_n \rangle$ is not uniformly convergent on $[0,1]$.

• Do you mean $f_n(x)=\frac{x}{1+nx^2}$ ? May 30, 2017 at 4:10
• or $f_n(x)=\frac{x}{1+nx^2}$? Just tell yes or no, I will edit for you? May 30, 2017 at 4:11
• I had Edited it. I don't know latex much. But I think done perfectly. May 30, 2017 at 4:13

Hint: A supremum on a compact interval is attained, so try to find the maximum of $f_n$ on $[0,1]$ by finding critical points of $f_n$ (set $f_n'(x) =0$). then you can accurately compute $M_n = \sup | f_n(x) - f(x)|.$
• First, as you said $K$ depends on $x$ and the inequality you wrote in the first line of your proof doesn't help you to get uniform convergence. you didn't use the theorem you said. you must compute $M_n$ first which is a number independent from $x$. May 30, 2017 at 4:27
• Definition is Definition is not a method of solving problem. As I said your way of solving this problem, what you wrote doesn't satisfy the Definition of uniform convergence, in definition $K$ must not depend on $x.$ And this line is wrong "$|f_n(x) - f(x)| < \varepsilon$ iff $1/nx < \varepsilon$ " May 30, 2017 at 4:38
• @AkashPatalwanshi: Because $A<B$ and $A<C$ do NOT imply that $B<C$. So your statement that "hence $|f_n(x)−f(x)|<\varepsilon$ if and only if $1/nx<\varepsilon$" is logically completely wrong. May 30, 2017 at 4:39