Real Analysis: Showing uniform convergence on $[, \infty)$ and $[0,\infty)$ for sequence of functions

Question

Let $f_n(x) = \frac{nx}{1+nx}$ for $ x \ge 0$.

I successfully proved that if $t > 0$, then convergence is uniform on $[t,\infty)$.

I am now asked to prove that convergence is not uniform on $[0,\infty)$.

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$\lim_{n \to \infty} f_n(x) = 0$ , $x = 0$

The problem is I can't believe I can use that in writing $|f_n(x) - f(x)|$ since $x=0$ is just one case. Any advice? Thank you greatly in advance!

Answer 1

Note that if $x\ne 0$, then $$ \lim _{n\to \infty } f_n(x) =\lim _{n\to \infty } \frac{nx}{1+nx}=1$$

On the other hand if $x=0$ we have $$ \lim _{n\to \infty } f_n(x) =\lim _{n\to \infty } 0 =0$$

Let $\epsilon =1/2$, then for a given $\delta >0 $ you can find two points, $x=\delta /2$ and $y=0$ such that $$ |x-y|< \delta $$ and $$|f_n(x)-f_n(y)|>1/2$$ for large enough n.

Thus the convergence is not uniform.

Answer 2

Assume that it were, then $f_{n}(x)\rightarrow\chi_{(0,\infty)}(x)$, we have for some $N$ that \begin{align*} \left|\chi_{(0,\infty)}(x)-\dfrac{nx}{1+nx}\right|<\dfrac{1}{2},~~~~n\geq N,~~~~x\in[0,\infty). \end{align*} Fix the $n=N$, then for all $x\in(0,\infty)$ we have
\begin{align*} \left|1-\dfrac{Nx}{1+Nx}\right|<\dfrac{1}{2}, \end{align*} taking $x\rightarrow 0^{+}$, we have $1<1/2$, which is absurb.

Answer 3

Prove by contradiction. Suppose the contrary that there exists a function $f:[0,\infty)\rightarrow\mathbb{R}$ such that $f_{n}\rightarrow f$ uniformly on $[0,\infty)$. In particular, $f_{n}\rightarrow f$ pointwisely. By direct calculation of pointwise limit, we have $$ f(x)=\begin{cases} 0, & \mbox{ if }x=0\\ 1, & \mbox{ if }x\in(0,\infty) \end{cases}. $$ Recall that uniform limit of continuous functions is also continuous, so $f$ is continuous, which is a contradiction.

Alternatively, we can prove it directly without using the fact that ``uniform limit of continuous functions is also continuous''. Since $f_{n}\rightarrow f$ uniformly, for $\varepsilon=0.01$, there exists $N\in\mathbb{N}$ such that $|f_{n}(x)-f(x)|<\varepsilon$ whenever $x\in[0,\infty)$ and $n\geq N$. In particular, $|f_{N}(\frac{1}{N})-f(\frac{1}{N})|<\varepsilon$. Note that $f_{N}(\frac{1}{N})=\frac{1}{2}$ while $f(\frac{1}{N})=0$. Therefore $|\frac{1}{2}-0|<0.01$ which is a contradiction.