# Checking Uniform Convergence

Question:

Prove that the sequence of functions $f_n(x) = x^n$ converges uniformly to zero on any interval of the form $[0,\mu]$ if $\mu < 1$.

My Work:

Recalling the definition of uniform convergence, we say that $f_n(x) \to 0$ uniformly if $$\forall\epsilon>0 \ \exists N \ \forall n \ge N \ \forall x \in \mathrm{Dom}\,(f) \ \Big[\left|f_n(x) - 0 \right|\le \epsilon \Big]$$

Since $x \in [0,\mu]$, we have $x \le \mu \Rightarrow \left| f_n(x) -f(x) \right| \le \mu^N$. We want $\mu^N \le \epsilon$ for some large $N$, so taking the natural logarithm of the inequality, which we can do because the natural log is monotone increasing for $x >0$, we have $$\mu^N \le \epsilon \Rightarrow N \ln \mu \le \ln \epsilon$$ Since $\mu, \epsilon < 0$ for $\epsilon$ sufficiently small, $\ln \mu, \ln \epsilon < 0$. So the inequality becomes $$N \ge {\ln \epsilon \over \ln \mu}$$

This makes sense to write because ${\ln \epsilon \over \ln \mu}$ is a positive number, so we can always take $N \ge {\ln \epsilon \over \ln \mu}$.

My Question: How do I incorporate this $N$ into an appropriately worded proof? I know it would go along the lines of "Taking $N \ge {\ln \epsilon \over \ln \mu}$, we have $$\mu^N - 0 \le \mu^{{\ln \epsilon \over \ln \mu}}$$ By virtue of my previous calculation, I know this should be less than $\epsilon$, but I am not sure how to perform the necessary algebra. Its a small, but crucial detail.

Edit:

Thanks to Nate Eldrege, who pointed out that $a^b = \exp(b \ln a)$. So we have $$\mu^{{\ln \epsilon \over \ln \mu}} = \mu ^{\ln(\epsilon - \mu)} = e^{\ln(\epsilon - \mu)}e^{\ln \mu} = \epsilon$$ So as required for any $n \ge N$, where the value of $N$ is given above, we have $\mu^N - 0 \le \epsilon$.

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Hint: $a^b = \exp(b \ln a)$. – Nate Eldredge Aug 23 '12 at 13:03
Remember that you solved $\mu^N = \epsilon$ to get this form of $N$. You should be able to go back again. – Sean Eberhard Aug 23 '12 at 13:05
I'd show uniform convergence differently: $f_n \to f$ uniformly means that $\|f-f_n\|_\infty \to 0$ means that $$\max_{x \in [0, \mu]} |f_n(x) - f(x)| \to 0$$ In this case, since $f=0$, this means that $$\max_{x \in [0, \mu]} |f_n(x) | \xrightarrow{n \to \infty} 0$$ To show this you need to find the extremal points of $f_n$. Since there are none with zero slope and since you know what $x^n$ looks like you see that there is one maximum on $[0, \mu]$ at $\mu$. – Rudy the Reindeer Aug 23 '12 at 13:05
(continued) But then $\|f_n\|_\infty = \max_{x \in [0, \mu]} |f_n(x) | = \mu^n \xrightarrow{n \to \infty} 0$ since $\mu \in [0,1)$ which shows the claim. – Rudy the Reindeer Aug 23 '12 at 13:06
@Matt xkcd.com/105 – Moderat Aug 23 '12 at 15:00

Nate Eldredge: "Hint: $a^b = \exp(b \ln a)$"