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