Uniform convergence of the sequence, $f_n:[a \geq 0,\infty) \to \mathbb R$ defined by $f_n(x)=\frac{n^2x}{1+n^3x^2},~n \in \mathbb N,$

Question

Let $f_n:[a \geq 0,\infty) \to \mathbb R$ be a sequence of functions defined by $$f_n(x)=\frac{n^2x}{1+n^3x^2},~n \in \mathbb N,$$ I found that, each $f_n$ has critical point $x_c=\frac{1}{n^{3/2}}$ and the value $f_n(x_c)=\frac{\sqrt n}{2}$ which is not going to $0$. While, I am not getting the monotonicity of each $f_n$ on $[a,\infty)$, can I say $f_n$ doesnot converge to $f \equiv 0$ uniformely on the same?

Answer 1

You can conclude that this funtion is not uniformly convergent on $[0,\infty)$.

Because $\frac{df_{n}(x)}{dx}=\frac{n^{2}}{1+n^{3}x^{2}}-\frac{2n^{5}x^{2}}{(1+n^{3}x^{2})^{2}}$.

So $$\frac{df_{n}(x)}{dx}\bigg |_{x=0}=\lim_{x\to 0^{+}}\frac{df_{n}(x)}{dx}=n^{2}$$.

But $f_{n}(x)\to 0\,,\forall x\in[0,\infty)$.

So $\frac{d}{dx}\lim_{n\to\infty}f_{n}(x)=\frac{d}{dx}(0)=0,\forall x\in[0,\infty)$.

But you can conclude uniform convergence in $[a,\infty)$ for $a>0$.

$$|f_{n}(x)|\leq |\frac{n^{2}x}{n^{3}x^{2}}|\leq \frac{1}{na}\forall,x\in[a,\infty)$$.

Thus it converges uniformly to $0$.