Let $\mathbb{K}=\mathbb{R}$, and let's consider, in $l^{2}$, the set $T=\left\{(\lambda_{n})_{n}\in l^{2}: \displaystyle\sum_{n=1}^{\infty}\frac{\lambda_{n}}{\sqrt{n}}=0\right\}$; I am asked to find $T^{\perp}$:
Here is my attempt: I am going to try to show that $T^{\perp}=\{\mathbf{0}\}$; for showing this, I am going to take $x=(\lambda_{n})_{n}\in T^{\perp}$, and try to show that $\lambda_{n}=0$, $\forall n\in\mathbb{N}$.
So, let, $x=(\lambda_{n})_{n}\in T^{\perp}$, then, by definition of $T^{\perp}$, $\forall y\in T$, $\langle x,y \rangle = 0$. I have seen that for example, $y_{1}=(1,-\sqrt{2},0,...,0,...)=e_{1}-\sqrt{2} \cdot e_{2}$ is an element of $T$, so taking on account what I have said, $\langle x,y_{1}\rangle =\lambda_{1}-\sqrt{2}\cdot \lambda_{2}=0$, and so $\lambda_{1}=\sqrt{2}\cdot \lambda_{2}$. Similiarly, is easy to show that $y_{n}=e_{1}-\sqrt{n}\cdot e_{n}$ is actually an element of $T$, so, we can deduce that $\lambda_{1}=\sqrt{n}\cdot \lambda_{n}$, $\forall n\in \mathbb{N}$... Nevertheless, I am not able to continue (in order to prove that $\lambda_{n}=0, \forall n\in \mathbb{N}$)... Someone could give me a hint/some help?
Thanks in advanced!
If $\lambda_1\ne0$, then $\lambda_n=\frac{\lambda_1}{\sqrt n}$. But $\left(\frac{\lambda_1}{\sqrt n}\right)_{n\in\Bbb N}=\lambda_1\left(\frac1{\sqrt n}\right)_{n\in\Bbb N}\notin\ell^2$. So, $T^\perp=\{0\}$.