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Consider the sequence of functions $f_n(x)=x^{1/n}$ Does this sequence converge uniformly or only pointwise on $[0,1]$? What about on $[1,2]$?

I know it converges pointwise on both intervals but I'm not sure whether the convergence is uniform.

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    $\begingroup$ The limit function is not continuous so the convergence is not uniform. $\endgroup$ – Kushal Bhuyan Dec 13 '16 at 3:18
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    $\begingroup$ Why Weierstass M? That's for series. $\endgroup$ – zhw. Dec 13 '16 at 3:38
  • $\begingroup$ @zhw. You are correct. The Weierstrass M test does not apply here. What should I use instead? $\endgroup$ – mathqueen459 Dec 13 '16 at 3:42
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For completeness, here is a proof that the convergence is uniform on $[1,2]$. Observe that on this interval, we have $$1 \leq x \leq 2$$ Since $f_n(x) = x^{1/n}$ is a monotonically increasing function on $[0, \infty)$, the above chain of inequalities is equivalent to $$1 = 1^{1/n} \leq x^{1/n} \leq 2^{1/n}$$ Subtracting $1$ throughout gives us $$0 \leq x^{1/n} - 1 \leq 2^{1/n} - 1$$ As a result, for all $x \in [1,2]$ we have $$|x^{1/n} - 1| \leq 2^{1/n} - 1$$ and therefore $$\sup_{x \in [1,2]}|x^{1/n} - 1| < 2^{1/n} - 1$$ Since the right-hand side converges to zero as $n \to \infty$, so must the left hand side. Thus, on $[1,2]$ we conclude that $f_n(x) = x^{1/n}$ converges to $1$ in the uniform norm, which is equivalent to uniform convergence.


Regarding uniform convergence on $[0,1]$, previous comments and answers show that this cannot be the case since $f_n(x) = x^{1/n}$ is continuous but the limit function $$f(x) = \begin{cases} 0 & \text{if }x = 0 \\ 1 & \text{if }x > 0 \\ \end{cases}$$ is not continuous. One might wonder whether removing the point $x=0$ from the domain would suffice to make the convergence uniform on $(0,1]$. This is not the case, and we can prove it using a similar argument to the one above. Specifically, note that for fixed $n$ we have $\inf_{x \in (0,1]}x^{1/n} = 0$, and $x^{1/n} \leq 1$ for all $x \in (0,1]$. Consequently, $$\begin{aligned} \sup_{x \in (0,1]} |x^{1/n} - 1| &= \sup_{x \in (0,1]} (1 - x^{1/n}) \\ &= 1 - \inf_{x \in (0,1]} x^{1/n} \\ &= 1 - 0 \\ &= 1 \\ \end{aligned}$$ so $x^{1/n}$ does not converge to $1$ in the uniform norm on $(0,1]$, even though it converges pointwise to $1$ on that interval.

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It does converge pointwise on $[0,1],[1,2]$ and $[0,+\infty[$, in fact. Defining $x^{\frac{1}{n}}$ as $0$ at $0$ and $e^{\frac{1}{n}log(x)}$ for $x>0$, one has

  1. $lim_{n\rightarrow +\infty}x^{\frac{1}{n}}=0$ for $x=0$
  2. $lim_{n\rightarrow +\infty}x^{\frac{1}{n}}=1$ for $x>0$

so, as the uniform limit of continuous functions is continuous, the convergence cannot be uniform on $[0,1]$ it is so however on $[1,2]$.

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For functions $f, g$ from any set $D$ to $\mathbb R$ let $\|f-g\|=\min (1,\sup \{|f(x)-g(x)|:x\in D\}).$

The def'n of uniform convergence on $D$ of the sequence $(f_n)_n$ to $f$ is that $\lim_{n\to \infty}\|f-f_n\|=0.$

When $D=[0,1]$ and $f_n(x)=x^{1/n}$ the sequence $(f_n(x))_n$ converges to $f(x)=1$ when $x\ne 0,$ and converges to $f(0)=0$ when $x=0.$ We have $$\|f-f_n\|\geq \min (1,|f(1/n^n)-f_n(1/n^n|)=\min (1,|1-1/n|)=1-1/n$$ so $(\|f-f_n\|)_n$ does not converge to $0.$

Theorem: When $D\subset \mathbb R$ and each $f_n:D\to \mathbb R$ is continuous on $D,$ and $(f_n)_n$ converges uniformly on $D$ to $f,$ then $f$ is continuous on $D.$

In the case above, each $f_n(x)=x^{1/n}$ is continuous on $D=[0,1]$ but $f$ is not continuous on $D,$ so the convergence can't be uniform.

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