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.
Thanks in advance!
Best Answer
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).