Find a sequence which converges in $X$, but not in $Y$, for:
- a) $X = \ell^\infty, Y = \ell^2$
- b) $X = \ell^2, Y = \ell^1$
This problem was left as an exercise to a reader and it comes with hints.
a) Since $\ell^\infty$ is a pretty wide space, I tried finding a sequence $x^n$ such that it isn't bounded in $\ell^2$ norm, but each element of the sequence is. So I came up with this sequence where term $\frac1n$ shows $n^3$ times:
$$x^n = (\frac1n, \frac1n, ... ,\frac1n, 0, 0, ...).$$
It is simple to see that $x^n$ converges to $0$ in $\ell^\infty$ as $$\|x^n\|_\infty = \frac1n \rightarrow 0, n \rightarrow \infty.$$
On the other hand, $x^n$ is not convergent in $\ell^2$ as it is not bounded in $\ell^2$ because of:
$$\|x^n\|_2^2 = n^3 \frac1{n^2} = n.$$
b) It seems impossible to use the same argument as in a). The hint for both of the problems state that I should find a sequence with norm equal to $1$ in $Y$. Thus I can use:
a) term $\frac1n$ shows $n^2$ times
b) term $\frac1n$ shows $n$ times
$$x^n = (\frac1n, \frac1n, ..., \frac1n, 0, 0, ...).$$
Now for both a) and b)
- in $X$ we have $\|x^n\| = \frac1n \rightarrow 0, n \rightarrow \infty$
- in $Y$ we have $\|x^n\| = 1$.
My questions are:
Why wouldn't such a sequence $x^n$ with property $\|x^n\| = 1$ be convergent (to a non-zero point)?"
Does it generally imply that a sequence with $\|x^n\| \ge 1$ diverges?
What if a sequence of linear bounded operators $A_n$ has $\|A_n\| \ge 1$?
Thank you for your help and time.