Uniform convergence of $\frac{n^3x}{1+n^4x^2}$

Question

I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.

So far i have attempted to prove pointwise converge as such;

Let $ N = 1/\epsilon x $

For every $n \ge N$ we have $ \vert\frac{n^3x}{1+n^4x^2} - 0 \vert = \frac{n^3x}{1+n^4x^2} < \frac{1}{Nx} = \frac{1}{(1/\epsilon x)x} = \epsilon $

Is this enough of a proof for pointwise convergence? and if so, how do i prove that the convergence is uniform (if it is in this case)?

Answer 1

Let $ n>0$.

For $ x=0 $, $ f_n(0)=0$, so $$\lim_{n\to+\infty}f_n(0)=0=f(0)$$

For $ x\in(0,1]$,

$$f_n(x)\sim \frac{n^3x}{n^4x^2}\sim \frac{1}{nx}$$ So, $$\lim_{n\to+\infty}f_n(x)=\lim_{n\to+\infty}\frac{1}{nx}=0=f(x)$$

The convergence is pointwise to zero function at $ [0,1]$.

For the uniform convergence, let $$g_n(x)=|f_n(x)-0|=\frac{n^3x}{1+n^4x^2}$$

the numerator of the derivative $ g'_n(x)$ is $$n^3(1+n^4x^2)-2n^4x.n^3x=$$ $$n^3(1-n^4x^2)$$ Thus

$$M_n=\sup_{x\in[0,1]}g_n(x)=g_n(\frac{1}{n^2})=\frac n2$$

But $\lim_{n\to+\infty}M_n=+\infty$, then the convergence is Not Uniform at $[0,1]$.

Answer 2

One strategy to check for uniform convergence is to find the maximum $M_n$ of $f_n(x) = \frac{n^3x}{1 + n^4 x^2}$ over $[0,1]$ using calculus, then check whether $M_n \to 0$ (keeping in mind that $0 < f_n(x) \leq M_n$ for all $x$).

That said, the same idea works with any upper bound for $f_n(x)$ that approaches $0$ as $n\to \infty$ and does not depend on $x$. For instance, one upper bound can be attained by combining an upper bound over $[0,1/n^2]$ and another over $[1/n^2,1]$.

With the first approach we find that $$ f_n'(x) = \frac{n^3(1 + n^4x^2) - n^3x(2n^4x)}{[1 + n^4 x^2]} = 0 \implies\\ n^3(1 + n^4 x^2) = 2 n^3 \cdot n^4 x^2 \implies\\ (1 + n^4 x^2) = 2 n^4 x^2 \implies\\ 1 = n^4 x^2 \implies\\ x = \pm 1/n^2. $$ We can see that the critical point at $x = 1/n^2$ is indeed a maximum since $f_n'(0) > 0$ and $f_n'(1) < 0$. Thus, our maximum is $$ M_n = f_n(1/n^2) = \frac{n^3 \cdot \frac 1{n^2}}{1 + n^4 \cdot \frac 1{n^4}} = \frac n2. $$ So, we find in fact that $M_n$ does not converge to $0$, which means that $f_n$ fails to converge uniformly.

Answer 3

Since $1/n \in [0,1]$,

$$\sup_{x \in [0,1]}\left|\frac{n^3x}{1+n^4x^2} \right| \geqslant \frac{n^3(1/n)}{1+n^4(1/n)^2} = \frac{n^2}{1+n^2} \underset{n \to \infty}{\longrightarrow} 1 \neq 0,$$

the convergence is not uniform.

Answer 4

Just note that for $(x_n)=1/n,\ \frac{n^3x_n}{1+n^4x_n^2}=\frac{n^2}{1+n^2}\to 1\neq 0$ as $n\to \infty$ so the convergence is not uniform.