I have a little struggle to show pointwise and uniform convergence of $ g_n (x) = \sqrt[n] {x} $ in the intervals of $(0, \infty), (0,1), [1, \infty) $

For $lim_{n \rightarrow \inf} g_n(x) = \begin{cases} 0 , for x=0 \\ 1,for 0 < x \leq 1 \end{cases} $

So in all those intervals the function does converge pointwisely.

Now my attempt for $(0,1)$ for uniform convergence: $| g_n(x) - g(x) |= | \sqrt[n] {x} - 1| \leq \epsilon$

In the next step I converted the inequation to n : $ \frac{ln(x)} {ln( \epsilon +1)} \geq n$

What does it tell me? And how do I proof uniform convergence in $(0, \infty), [1, \infty) $? Any help really appreciated.


1 Answer 1


Since each $g_n$ is unbounded on $(0,+\infty)$ and the constant function $1$ is bounded, the convegence cannot be uniform. In fact, if $(g_n)_{n\in\mathbb N}$ converged uniformly to $1$, then there would be a natural number $p$ such that$$(\forall n\in\mathbb{N})\bigl(\forall x\in(0,+\infty)\bigr):n\geqslant p\implies\bigl|1-\sqrt[n]x\bigr|<1\implies\sqrt[n]x\in(0,2);$$just use the definition of uniform convergence with $\varepsilon=1$.

Of course, the same argument applies to $[1,+\infty)$.

  • $\begingroup$ Thank you, that's what I suspected too.. But how can I formally proof it? $\endgroup$ Nov 26, 2017 at 10:52
  • $\begingroup$ @wondering1123 I've edited my answer. Is it clear now? $\endgroup$ Nov 26, 2017 at 10:58

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