-1
$\begingroup$

Let $K$ be the subset of $\ell^p$, $K \subset \ell^p$ by $$K = \{ k_x + xk_y \ \vert \ x < y, x,y\}.$$

Note that $x$ and $y$ are assumed to be natural numbers. And $k_x$ is the sequence consisting of $0$ in the $i$th place where $i \neq x$ and 1 in the $x$th place.

Show that $K$ is closed in the strong topology and deduce that $0$ is in the closure of $K$ with respect to the weak topology.

My attempt so far: Consider some sequence $(a_n) \subset K$ such that $a_n$ converges to some $a \in \ell^p$. Since we want to show strong convergence, we have that for $\epsilon >0$ some $N$ such that $n \geq N$ implies $$\| a_n - a \|_p = \left( \sum_{n=1}^y \left| a_n -a \right|^p \right)^{1/p} = \left( \sum_{n =1}^y \left| k_n + nk_y \right| \right) < \epsilon$$ Note that I've defined the sequence $(a_n) = \{ k_n + nk_y | n < y\}$.

I now need to prove that $a = k_x + xk_y$ for some $x

  • 0
    @Did May be able to help you out. – 2017-02-02
  • 1
    What is $k_x$? Is it the indicator sequence of $x$? Also, recall that a strongly convergent sequence is also strongly Cauchy. – 2017-02-02
  • 0
    Furthermore, are there any limitations on $p$ we should be aware of? – 2017-02-02
  • 0
    @Aweygan $k_x$ is the sequence consisting of $0$ in the $i$th place where $i \neq x$ and 1 in the $x$th place. – 2017-02-02
  • 1
    @user3359: Your stated equation is wrong. Note that $n$ is independent variable (with $n \ge N$) on the left-hand side, but after that it is the summation index. – 2017-02-02

1 Answers 1

0

HINT You can show more than $a\in K$, you can show that the sequence is constant after some time. Since $a_n\to a$, the sequence $\{a_n\}$ is Cauchy, hence there is some $N\in\mathbb N$ such that $\|a_n-a_m\|_p<1$ for all $n,m\geq N$. This tells us that $a_N=a_n$ for all $n\geq N$ (why?), and therefore $a=a_N\in K$.

  • 0
    Thanks for your response, with regard to your second hint, I don't think that converges weakly to zero. The weak topology is not metrizable and so there is no sequence in $K$ that converges weakly to zero. Can you explain your hint more perhaps, or resolve this apparent discrepancy. – 2017-02-04
  • 0
    @user3359 after posting this, I have been second guessing myself. I'll edit to remove that suggestion. Instead, try showing that any weak neighborhood of $0$ contains an element of $K$. – 2017-02-04
  • 0
    @user3359 I don't understand your reasoning when you say the weak topology is not metrizable implies no sequence in $K$ converges to $0$. – 2017-02-04
  • 0
    Nevermind, I've solved the problem. Thanks for your help. – 2017-02-04