Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mu^N - 0 \le \epsilon$.

share|improve this question
1  
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
1  
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 $\mu$. –  Matt N. 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. –  Matt N. Aug 23 '12 at 13:06
1  
@Matt xkcd.com/105 –  Mike Aug 23 '12 at 15:00

1 Answer 1

up vote 0 down vote accepted

Seeing as I have found my answer, and this question needs no more work, I am posting Nate Eldredge's hint as an answer and accepting it so that this question moves off the "Unanswered" list:

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

share|improve this answer
    
But it will only let me accept my own answer in 2 days as I have just found out, so if someone posts an answer before then I'll accept it. –  Mike Aug 23 '12 at 13:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.