Let $H$ be some Hilbert space with an orthonormal basis $\{e_n\}$. Set $\lambda = (\lambda_n) \in \ell^\infty$ and define $M_\lambda:H \rightarrow H$ by $M_\lambda x = \sum_{n=1}^\infty \lambda_n\langle x,e_n\rangle e_n$. Show that $M_\lambda$ is surjective if and only if $\inf |\lambda_n| > 0$.
One of the directions is simple enough, if $\inf|\lambda_n| > 0$, then $|\lambda_n| > 0$ for every $n$. So if $x \in H$, then $$\sum_{n=1}^\infty \left|\frac{1}{\lambda_n} \langle x, e_n \rangle \right|^2 < \frac{1}{\delta^2} \sum_{n=1}^\infty |\langle x,e_n\rangle|^2 < \infty.$$ So the series defined by $z = \sum_{n=1}^\infty \frac{1}{\lambda_n} \langle x,e_n \rangle e_n$ converges and it is clear that $M_\lambda(z)=x$.
What I am having trouble with is the other direction. I think this is done by contradiction, assuming that $\inf |\lambda_n| = 0$ and showing that $M_\lambda$ is in fact not surjective. However, I am unsure how to construct a candidate $x$ so that $M_\lambda(z) \neq x$ for any $z$. Please help. Thanks!
Best Answer
If the infimum is $0$ then we can find a subsequence $(\lambda_{n_k})$ such that $|\lambda_{n_k}| <\frac 1 {2^{k}}$ for all $k$. Let $a_n=0$ if $n$ is not of the form $n_k$ and let $a_{n_k}=\frac 1 k$. The $x=\sum a_ne_n \in H$ and if $M_{\lambda} y=x$ with $y=\sum y_ne_n$ the we have $\lambda _n y_n=a_n$ for all $n$. But $\|y\|^{2}=\sum |y_n|^{2} \geq \sum \frac {2^{2k}} {k^{2}}=\infty$ a contradiction.