Pointwise and Uniform Convergence on a specific intervall

Question

I have the sequence of functions $$ f_n(x) = \frac{x}{x^2+ \frac{1}{n}} \quad x \in [0, \infty)$$ and I need to show that the sequence converges pointwise as well as uniformly, however only on the intervall $[\alpha, \infty) $ where $\alpha > 0 $.

I started out finding the limit function of $f_n(x)$ in the following manner:

$$ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{x}{x^2+ \frac{1}{n}} \xrightarrow{n \to \infty} \frac{x}{x^2} = \frac{1}{x} $$ which leads me to my first question: Is this enough to show that $f_n(x)$ converges pointwise to $\frac{1}{x}$ ?

I have at first gone on with the idea that I can actually do that, which leads then to my formula for uniform convergence:

$$ |f_n(x) - f(x)| = \left|\frac{x}{x^2 + \frac{1}{n}} - \frac{1}{x}\right| = \left| \frac{-1}{nx^3 + x}\right| \leq \frac{1}{nx^3}$$

My second question is of course if this is at all allowed. If it is, is it then enough to show that as long as $x > 0$ this function is defined and therefore this function is uniformly convergent?

Answer 1

You proved correctly that $\lim_{n\to\infty}f_n(x)=\frac1x$ if $x>0$. You should add to this that $\lim_{n\to\infty}f_n(0)=0$. Therefore, your sequence converges pointwise to the function $f\colon[0,+\infty)\longrightarrow\mathbb R$ defined by$$f(x)=\begin{cases}\frac1x&\text{ if }x>0\\0&\text{ if }x=0.\end{cases}$$So, the convergence cannot be uniform, since the uniform limit of a sequence of continuous functions is always continuous.

In order to prove that it converges uniformly on $[\alpha,+\infty)$, you can note that$$\left|\frac x{x^2+\frac1n}-\frac1x\right|=\frac1n\left|\frac1{x^3+\frac xn}\right|=\frac1{nx^3+x}\leqslant\frac1{n\alpha^3+\alpha}\to0.$$

Answer 2

Uniformly in $[\alpha,\infty],$ $\alpha >0$

You have :

$|f_n(x)-f(x)| \lt \dfrac {1}{nx^3} \le \dfrac{1}{n\alpha^3}.$

Let $\epsilon >0$ be given.

Archimedes' principle:

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_0 \gt (\epsilon\alpha^3)^{-1}.$

For $n \ge n_0$ we have

$|f_n(x)-f(x)| \lt \dfrac {1}{n\alpha^3} \le \dfrac{1}{n_0\alpha^3} \lt \epsilon.$

Answer 3

All you did so far is correct. It remains to show that the convergence $f_n \to \frac1x$ is uniform on $[\alpha, +\infty)$:

$$\left|f_n\left(x\right) - f\left(x\right)\right| = \frac1{nx^3+x} \le \frac{1}{nx^3} \le \frac1{n\alpha^3} \xrightarrow{n\to\infty} 0$$ uniformly in $x \in [\alpha, +\infty)$.

The convergence $f_n \to \frac1x$ is not uniform even on $(0, +\infty)$:

$$\left|f_n\left(\frac1n\right) - f\left(\frac1n\right)\right| = \frac{1}{n\left(\frac1n\right)^3 + \frac1n} = \frac{n^2}{n+1}$$ which does not converge to $0$ as $n\to\infty$.