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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 $[0,1]$.

May be I am wrong somewhere in last proof. Please help me.

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

1 Answer 1

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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 $M_n = \sup | f_n(x) - f(x)|.$

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  • $\begingroup$ I need to know why I am Not getting uniform convergence of given sequence, if I go through definition of uniform convergence of sequence. Is I am wrong somewhere? (In inequalities?) $\endgroup$ May 30, 2017 at 4:19
  • $\begingroup$ 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 $x$. $\endgroup$
    – Red shoes
    May 30, 2017 at 4:27
  • $\begingroup$ I meant, if we go via theorem, we get uniform convergence of sequence(I known) "But if we go via definition then why I am not getting uniform convergence of sequence? $\endgroup$ May 30, 2017 at 4:29
  • $\begingroup$ 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 $1/nx < \varepsilon$ " $\endgroup$
    – Red shoes
    May 30, 2017 at 4:38
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    $\begingroup$ @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. $\endgroup$
    – zipirovich
    May 30, 2017 at 4:39

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