# Show that $f_n \rightarrow 0$ in $C([0, 1], \mathbb{R})$

I was given the following problem and was wondering if I was on the right track.

Let $$f_n(x) = \frac{1}{n} \frac{nx}{1 + nx}, \: 0 \le x \le 1$$

Show that $$f_n \rightarrow 0$$ in $$C([0, 1], \mathbb{R})$$.

I have this theorem that I figured I could use:

$$f_k \rightarrow f$$ uniformly on A $$\iff$$ $$f_k \rightarrow f$$ in $$C_b$$.

In this case, $$C_b$$ is the collection of all continuous functions on $$[0,1]$$. So if I can prove the function is uniformly continuous, this would prove that $$f_n \rightarrow 0$$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?

Thanks

• The M test is useless here because it's not a series. In any case, you should be able to simply compute $\| f_n \|_\infty$... – Ian Dec 7 '18 at 1:07

note that for $$x \in [0,1]$$ we have $$|f_n(x)| =|\frac{1}{n}\frac{x}{1+nx}| \le \frac{1}{n}$$, so $$||f_n||_\infty \le \frac{1}{n}$$ which implies $$f_n \to 0$$ in $$C_b[0,1]$$
• So are you saying I can prove it converges in $C_b$ directly by the method you showed? I don't need to show uniform convergence? – user591271 Dec 7 '18 at 1:52
• Yes. However the inequality shown above, stating that $|f(x)|\le \frac{1}{n}$ holds for all $x \in [0,1]$ directly renders uniform convergence as well. – Maksim Dec 7 '18 at 2:28
You'll notice that you can cancel to get $$f_n(x) = \frac{x}{1+nx}$$. For $$x =0$$, clearly $$f_n(x)= 0$$. Otherwise, $$nx$$ gets arbitrarily large. This will imply that $$f_n(x) \to 0$$.
• I find having the $n$'s there is actually useful, because it suffices to note that $f(y)=\frac{y}{1+y}$ is a bounded function on $[0,\infty)$, and then the $1/n$ outside creates the decay. – Ian Dec 7 '18 at 14:38