# Uniform convergence of the sequence of functions $f_n(x) = \frac{x}{1+n^2x^2}$ on the interval $[0,1]$

I've got another question relating to the uniform convergence of a sequence of functions (sorry). The sequence of functions in question is:

$f_n(x) = \displaystyle\frac{x}{1+n^2x^2}$ on the interval $[0,1]$

I've already shown that $h_n(x) = \displaystyle\frac{nx}{1+n^2x^2}$ converges, but not uniformly by looking at $x = \frac{1}{n}$

But for $f_n(x)$ clearly $\displaystyle\lim_{x \to +\infty} f_n(x) = f(x) = 0$ for every $x \in [0,1]$

So if $f_n(x)$ were uniformly convergent we would have $\bigg|\displaystyle\frac{x}{1+n^2x^2}\bigg| = \bigg|\frac{1}{\frac{1}{x} + n^2x}\bigg| < \epsilon \$ for some given $\epsilon > 0$, $n > N \in \mathbb{N}$ and $\forall x \in [0,1]$

Or, alternatively, by Cauchy's criterion we would have

$|f_n(x) - f_m(x)| \leq \bigg|\displaystyle\frac{1}{n^2x} - \frac{1}{m^2x}\bigg| < \epsilon$

For $m > n > N$ and $\forall x \in [0,1]$

The problem is, I'm not convinced if $f_n(x)$ is uniform or not, because while, given any $\epsilon$, we could find a smaller $x$ to increase the value of $\bigg|\frac{1}{n^2x}\bigg|$ this is counteracted by the $\frac{1}{x}$ term also on the denominator. So I'm not sure where to go in regards to proving whether this function is uniformly convergent or not.

• Can you find the maximum value of $f_n$ on $[0,1]$? The derivative is not hard to analyze. – David Mitra Feb 26 '13 at 12:18
• @David Hi David, this question is asked prior to covering derivatives in analysis so I wanted to answer it without using differentiation. Thanks for the suggestion though! – Noble. Feb 26 '13 at 12:19

Basically, there are two useful observations here, valid for all $x$:

$$f_n(x) < x \quad \text{and} \quad f_n(x) < \dfrac1{1+n^2x^2}$$

Let $\epsilon > 0$ be arbitrary. If $x < \epsilon$ then by the first estimate, $|f_n(x) - f(x)| = f_n(x) < \epsilon$. Next we observe that the second estimate $\dfrac1{1+n^2x^2}$ is monotonically decreasing for $x \in [0,1]$.

It is thus sufficient (together with our earlier observation) to pick an $N$ satisfying $\dfrac1{1+N^2\epsilon^2} < \epsilon$; the monotonicity then provides $\dfrac1{1+N^2x^2} < \epsilon$ for all $x \in [\epsilon, 1]$.

Hence $f_n$ is uniformly convergent on $[0,1]$. The method used is quite common: to use a different estimate close to one of the boundaries, thereby obtaining a new interval of uniform convergence that permits an easier assessment (for example because the bounds of the interval depend on $\epsilon$, as in this particular case).

• Thanks a lot for the helpful explanation. – Noble. Feb 26 '13 at 12:18
• You're welcome, HTH. – Lord_Farin Feb 26 '13 at 12:35

Fix $n$ and study the function $f_n$ on $[0,1]$.

We have $$f_n'(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}.$$

So $f_n$ is nonnegative and attains its maximum at $1/n$ on $[0,1]$.

Hence $$0\leq \sup_{[0,1]} f_n=f_n(1/n)=\frac{1}{n}.$$

By the squeeze theorem, it follows that $f_n$ converges uniformly to $f(x)=0$ on $[0,1]$.

• A small correction. $f_n(\frac{1}{n}) = \frac{1}{2n}$. – user154185 Jun 7 '14 at 16:31