# Solution to operator equation in Hilbert space

Suppose $H$ is a Hilbert space with orthonormal basis $\{e_n\}$. Suppose that $\{\alpha_n\}$ is a sequence of complex numbers with $\lim_{n\rightarrow\infty} \alpha_n = 0$, and define $T: H\rightarrow H$ by $Tx = \sum_{n=1}^\infty \alpha_n\langle x, e_n\rangle e_n$.

When does the equation $(T-\alpha I)z = x$ have a solution $z$?

By comparing coefficients we see that if $z = \sum \beta_n e_n$, then we need $$\beta_n = \frac{1}{\alpha_n -\alpha} \langle x, e_n \rangle.$$

So provided that $\langle x, e_n\rangle = 0$ whenever $\alpha_n = \alpha$ then we're fine. At least, this is what I got at first, and is also essentially what the answer here Equation in Hilbert space says. However, what happens if $\alpha = 0$? Then the sequence $\{\beta_n\}$ is unbounded since $\alpha_n\rightarrow 0$, and hence not square summable, so the definition of $z$ doesn't converge.

This would suggest to me that the equation, which is now just $Tz = x$, only has a solution if $\langle x, e_n\rangle \neq 0$ for only finitely many $n$, in order for $z$ to converge. But this means that $T$ is finite rank, but that's certainly not necessary because it could be the case that all the $\alpha_n$ are nonzero, in which case $T$ is still a well defined and bounded operator.

What am I missing here?

First of all even if $Tz = x$ has only solutions if $\langle x,e_n\rangle\neq 0$ for only finitely many $n$ then still every $e_n$ for which the corresponding $\alpha_n$ fulfills $\alpha_n \neq 0$ is in the range of $T$ since $T\frac{1}{\alpha_n} e_n = e_n$. Hence the $\dim\operatorname{ran} T$ is infinite if there is no $n_0\in\mathbb{N}$ such that $\alpha_n = 0$ for all $n\geq n_0$.
Anyway there is something you forgot: $\beta_n = \frac{1}{\alpha_n}\color{red}{\langle x,e_n\rangle}$ so $\beta_n$ must not be unbounded as $n\to\infty$ since the red marked $\langle x,e_n\rangle$ could shrink faster than $\alpha_n$.
for example take $\langle x,e_n\rangle = \frac{\alpha_n}{n}$ then we receive $\beta_n = \frac{1}{n}$ for all $n\in\mathbb{N}$. I hope this solves your confusion.
• So how do I ensure that my $z$ is a solution given the information that I have? – IAlreadyHaveAKey May 27 '17 at 14:16
• Suppose that $\langle x, e_n\rangle = 0$ whenever $\alpha_n = 0$, in the case where $\alpha = 0$. Does the equation have a solution? Or do we need more assumptions on $x$. – IAlreadyHaveAKey May 27 '17 at 14:24
• In this case you also need that $\frac{1}{\alpha_n} \langle x,e_n\rangle$ (this is well defined since we say $\frac{0}{0} =0$ for now) is quadratic summable. since we want $\sum_{n=1}^\infty \frac{1}{\alpha_n} \langle x,e_n\rangle e_n$ to converge. – Nathanael Skrepek May 27 '17 at 14:29