Problem
Suppose $V$ is a vector space and $T: V \rightarrow V$ is a linear map such that $\|Tv\| \leq \|v\|$ for every $v \in V$. Prove that $T-\sqrt{2}I$ is invertible.
The case that $V$ is a finite-dimensional is clear:
If $v \in N(T-\sqrt{2}I)$, then $Tv = \sqrt{2}v$, so $\|Tv\| = \sqrt{2}\|v\| \leq \|v\|$ implies $\|v\|=0$.
My question is, does this result remain true in the infinite dimensional case? That is to prove the surjectivity of $T-\sqrt{2}I$.
On
If you wanted to solve $(T-\sqrt{2}I)x=y$ for $x$, then would have $$ x = \frac{1}{\sqrt{2}}Tx-y \\ = \frac{1}{\sqrt{2}}T(\frac{1}{\sqrt{2}}Tx-y)-y \\ = (\frac{1}{\sqrt{2}}T)^2 x-\frac{1}{\sqrt{2}}Ty-y \\ = ... = (\frac{1}{\sqrt{2}}T)^{N+1}x-\sum_{n=0}^{N}(\frac{1}{\sqrt{2}}T)^ny $$ Because of the assumptions on $T$, if $x$ were a known solution, it would have to be true that the following series would converge in the norm of the vector space: $$ x = -\sum_{n=0}^{\infty}(\frac{1}{\sqrt{2}}T)^ny. $$ So, knowing $(T-\sqrt{2}I)x=y$ forces the right side to converge in norm to $x$, and the solution is seen to be uniquely given by the sum. Conversely, if the sum converges, then the assumptions on $T$ force the above to hold in the limit. But there is no guarantee that the sum will convergen in general. It must if $x$ is a known solution. However if the space is complete, meaning that norm Cauchy sequences converge in the space to a vector, then $(T-\sqrt{2}I)x=y$ has a unique solution $x$ for every $y$ that is given by the infinite sum, which is forced to converge by completeness.
If $S$ is any linear operator on some Banach space (i.e. a complete normed vector space) with $\|S\| < 1$ (from which it follows that $\|Sv\| < \|v\|$ for all $v$), then there is a fairly standard argument that $I-S$ is invertible. The basic idea is to look at the power series expansion of $(1-s)^{-1}$ (where $s$ is a real or complex variable), but substitute $S$ for $s$ and show that the same argument works. This standard argument is sketched below:
Note that if $|s| < 1$, then $$ \frac{1}{1-s} = \sum_{j=0}^{\infty} s^j. $$ It is this series that we are going to show does the job in a more general space. That is, this series should give the inverse of $I-S$ in a Banach space. Indeed, observe that if $\|S\| < 1$, then the series $$ \sum_{j=0}^{\infty} S^j $$ is Cauchy, and so by completeness converges to some operator (note that we really do require a Banach space here, and not just any old normed vector space). But then $$ (I-S) \left( \sum_{j=0}^{\infty} S^j \right) = \sum_{j=0}^{\infty} (S^j - S^{j+1}) = \sum_{j=0}^{\infty} S^j - \sum_{j=0} S^{j+1} = \sum_{j=0}^{\infty} S^j - \left( \sum_{j=0}^{\infty} S^{j} - I\right) = I. $$ Multiplication by $(I-S)$ on the other side gives the same result, from which it follows that $$ (I-S)^{-1} = \sum_{j=0}^{\infty} S^j. $$ In particular, $I-S$ is invertible whenever $\|S\|<1$.
In the case of the original question, assume that $\|T\| \le 1$ and define $S := \frac{1}{\sqrt{2}} T$. Since $\|T\| \le 1$, it follows that $$ \|S\| = \left\| \frac{1}{\sqrt{2}} T \right\| = \frac{1}{\sqrt{2}} \|T\| \le \frac{1}{\sqrt{2}} < 1. $$ Therefore we may apply the result at the top and conclude that $I-S$ is invertible, with inverse given by $$ (I-S)^{-1} = \sum_{j=0}^{\infty} S^j. $$ But then $$ I = (I-S) \sum_{j=0}^{S^j} = \left[ \sqrt{2}(I-S) \right] \left[ \frac{1}{\sqrt{2}} \sum_{j=0}^{\infty} S^j \right] = (\sqrt{2}I - T) \sum_{j=0}^{\infty} \frac{S^j}{\sqrt{2}}, $$ from which it is possible to conclude that $\sqrt{2}I - T$ is invertible, with inverse given by the series above.