Axler LADR, exercise 6.A-5 This question concerns exercise 6.A-5 in Axler's book. It is difficult to solve for an infinite-dimensional case. Maybe it can't be solved.
It all boils down to the following question:
Suppose we have an infinite-dimensional vector space. There is a linear map that for any vector $v$, sends $v$ to a vector, which norm is not greater than that of $v$. Will the linear map $T - \sqrt{2} I$ be surjective?
(I found in the errata for this book that we should assume that the space is finite-dimensional. But it still interesting to solve the problem.)
 A: This is the chapter in which Adler discusses inner product spaces and he assumes all his vector spaces $V$ are, in fact, inner product spaces. Recall that for any linear map $S\in \mathcal L(V)$ we have $(\text{im}\, S)^\perp \subset \ker S^*$, where $S^*$ is the adjoint map characterized by $\langle Sv,w\rangle = \langle v,S^*w\rangle$ for all $v,w\in V$.
Let $S=T-\sqrt2 I$. We claim that $\ker S^* = \{0\}$. For suppose $S^*w=0$. Then
$$0=\langle S^*w,w\rangle = \langle (T^*-\sqrt2I)w,w\rangle = \langle w,Tw\rangle - \sqrt2\|w\|^2. \qquad\tag{$\star$}$$
But, by Cauchy-Schwarz, $\langle w,Tw\rangle \le \|w\|\|Tw\|\le \|w\|^2$, and so the right-hand side of ($\star$) has norm at least $(\sqrt2-1)\|w\|^2$. It follows that $w=0$.
We deduce now that $\text{im}\, S$ is dense. However, since $\|S\|\ge \sqrt2-1$, it is known that the image of $S$ is closed if we assume $V$ is a Banach space (see this, for example). Thus $S$ is surjective.
COMMENT: Given that all the exercises near this one in the book are quite easy, I expect Adler intended $V$ to be finite-dimensional. We certainly aren't equipped to make all these additional assumptions, and I don't even think he's discussed the geometry of the adjoint at this point in the text.
