0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $[1,+\infty)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, that's what I suspected too.. But how can I formally proof it? $\endgroup$ – wondering1123 Nov 26 '17 at 10:52
  • $\begingroup$ @wondering1123 I've edited my answer. Is it clear now? $\endgroup$ – José Carlos Santos Nov 26 '17 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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