I'm a bit stuck with this question and I would appreciate any help.
Let's assume we have $\ell^1$ space of sequences over field $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$ with norm inherited from $\ell^2$ space. We define (linear) functional $T: l^1 \rightarrow \mathbb{K}$ by
$$T(x) = \sum_{n=1}^{\infty}\ \frac{x_n}{\sqrt{n}}.$$
I would like to check if this operator is bounded or unbounded. Now, I'm not sure if I approach this question correctly. I can show that this operator is unbounded on the space $\ell^2$ by considering sequence $x = (x_n) = \frac{1}{\sqrt{n}ln(n+1)}, n \in \mathbb{N}$, since then $x \in \ell^2$ and after normalizing this vector we will get that $T(x) = \infty$.
Nevertheless, if I understand the question correctly, we need to find such $(x_n) \subset \ell^1$ that $\|x_n\|_2 = 1$ and $T(x_n) \rightarrow \infty$ (so my previous example doesn't work since $x \notin \ell^1$).
Is it possible to find such a sequence? My initial intuition was that this function should be unbounded since I couldn't find a good bound for this operator (again, with respect to $\ell^2$ norm, since with respect to $\ell^1$ norm I would just bound it by $\|x\|_1$, and Hölder's inequality didn't work since $(\frac{1}{\sqrt{n}}) \notin \ell^2$) but perhaps there is a better bound for this operator?
Thanks!
Best Answer
You are on the right track, you noticed that for the sequence $x = \left(\frac1{\sqrt{n}\ln (n+1)}\right)_n \in \ell^2$, $Tx$ is not even well-defined.
Now we can approximate $x$ by vectors $(v_n)_n$ in $\ell^1$, namely set $$v_n := \left(\frac1{\sqrt{1}\ln 2}, \frac1{\sqrt{2}\ln 3}, \ldots, \frac1{\sqrt{n}\ln (n+1)}, 0, 0, \ldots\right).$$
We have $$\|v_n\|_2 = \sqrt{\sum_{k=1}^n\left(\frac1{\sqrt{k}\ln (k+1)}\right)^2} \le \underbrace{\sqrt{\sum_{k=1}^\infty \left(\frac1{\sqrt{k}\ln (k+1)}\right)^2}}_{M:=} <+\infty$$ but $$|Tv_n| = \sum_{k=1}^n \frac{1}{k\ln (k+1)} \xrightarrow{n\to\infty} \sum_{k=1}^\infty \frac{1}{k\ln (k+1)} = +\infty$$ so $T$ maps a bounded sequence $(v_n)_n$ in $\ell^1$ to an unbounded sequence in $\mathbb{K}$. Hence $T$ is unbounded.