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.

I'd like your help with the deciding whether the following function uniformly converges in two intervals.

$$f_n(x)=n \left(x^{\frac{1}{n}}-1 \right) .$$

In $(0,10]$, with l'Hôpital, I showed that for every $x$ the limit of the function is $\ln x$, so with a help from Dini's theorem, I am able that the function is monotone and the limit is continuous so the function uniformly converges.

I'm not sure what should I do with the other interval. I tried to evaluate the limit of $|f_n(x)-f(x)|$, I didn't come to any smart conclusion.

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

The sequence does converge uniformly to $\ln x$ on any compact subset of $(0,\infty)$, but the convergence is not uniform in $(0,a\,]$ for any $a>0$. If the convergence were uniform, there would exist an $n_0$ such that for all $n\ge n_0$ $$ |f_n(x)-\ln x|\le1\quad\forall x\in(0,a\,]. $$ But $$ |f_n(2^{-n})-\ln(2^{-n})|=\Bigl(\ln2-\frac12\Bigr)n $$ is not bounded.

Observe that $(0,a\,]$ is not compact and Dini's theorem does not apply.

share|improve this answer
add comment

As pointed out by Julian, the sequence does not converge uniformly on $(0,a]$ for any $a>0$; but it does converge uniformly on any compact subset of $\Bbb R$.

For sets of the form $[a,\infty)$:

Using L'Hopital, you can show the pointwise limit is $f(x)=\ln x$ on $[a,\infty)$.

Now take $x_n=2^n$. Then $$ |f_n(x_n)-\ln(x_n)|= |n(2-1) - n\ln 2|=n(1-\ln2)\quad \buildrel{n\rightarrow\infty}\over\longrightarrow\quad\infty. $$ Thus, there is no $N$ so that $|f_n(x)-\ln x|<1$ for all $n\ge N$ and all $x\in [a,\infty)$. Thus, the convergence is not uniform on $[a,\infty)$.

share|improve this answer
add comment

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.