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?


  • $\begingroup$ The M test is useless here because it's not a series. In any case, you should be able to simply compute $\| f_n \|_\infty$... $\endgroup$ – 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 $C_b[0,1]$

  • $\begingroup$ Nice job. It's exactly what I would put. Well done. $\endgroup$ – ncmathsadist Dec 7 '18 at 1:27
  • $\begingroup$ 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? $\endgroup$ – user591271 Dec 7 '18 at 1:52
  • $\begingroup$ 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. $\endgroup$ – 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 $f_n(x) \to 0$.

  • $\begingroup$ 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. $\endgroup$ – Ian Dec 7 '18 at 14:38

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