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Let $f_n(x)=(1+x^n)^\frac{1}{n}$ on $[0, \infty)$.

I want to check if this is uniformly convergent.

It's pointwise limit is $$f(x)=\begin{cases} x \text{ if } |x|\geq1\\ 1 \text{ if } |x|<1\\ \end{cases}$$

My first attempt was to use the Weierstrass M-test to check if its uniformly convergent. I have to do this for two cases $x<1$ and $x \geq 1$. Setting the derivative equal to zero:

$(x^n+1)^{\frac{1}{n}-1}x^{n-1}-1=0$
$(x^n+1)^{\frac{1}{n}-1}x^{n-1}=1$

At this point I get stuck. I tried using SageMath to no avail. I'm wondering if I am approaching this problem correctly.

emka
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1 Answers1

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First notice that $(1+x^n)^{1-\frac{1}{n}} \geq x^{n-1}$ so that $0\leq f_n'(x) \leq 1$ for all $x\in [0,\infty)$. This gives that, for $x\geq 1$, $f_n'(x)-f'(x)\leq 0$ meaning this difference is non-increasing so $$ \sup_{x\in [0,\infty)} |f_n(x)-f(x)| \leq f_n(1)-f(1) = 2^{1/n}-1 \to 0 $$ so the convergence is uniform in $[1,\infty)$. The bound for $f_n'$ and Arzelà-Ascoli give the uniform convergence in $[0,1]$.

Jose27
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  • I'm not sure I understand why $(1+x^n)^{1-\frac{1}{n}}\geq x^{n-1}$ shows that $0\leq f_n'(x)\leq 1$. – emka Mar 23 '13 at 18:37
  • Well $f_n'(x)=\frac{x^{n-1}}{(1+x^n)^{1-\frac{1}{n}}}$. – Jose27 Mar 23 '13 at 18:47
  • Thanks. I think my first few readings of your solution were quite haphazard as I thought you had somehow showed uniform convergence for all of $[0, \infty)$ simultaneously instead of splitting into cases of $x< 1$ and $x \geq 1$. – emka Mar 23 '13 at 19:26
  • Acutally, one more quick remark, how did you get the derivative of $f'_n(x)$ to be that fraction. I get the derivative as $(1+x^n)^{\frac{1}{n}-1}x^{n-1}$. Does it follow changing $\frac{1}{n}-1$ to $1-\frac{1}{n}$? Why wasn't $x^{n-1}$ changed to $\frac{x^n}{x}$? – emka Mar 23 '13 at 19:50
  • You're right, it follows from changing the exponent. For the second question I guess for convinience, the way I expressed the derivative works best (for me) when getting the desired bounds. – Jose27 Mar 23 '13 at 20:12