I recently picked up Peter J Olver's "Introduction to Partial Differential Equations" again (Read the full text here). I am on chapter 9, "A General Framework for Linear PDEs". The book is dealing with linear operators in fairly general terms. There is a certain theorem that I cannot understand, however. Page 355 in the linked text. Here, $U$ is a vector space over $\mathbb{R}$ equipped with an inner product $\langle\cdot,\cdot\rangle$ and $S\in\operatorname{End}(U)$ is positive definite, i.e $\langle u,Su\rangle>0~~\forall u\in U\setminus\{0\}$. Here is the proposition:
Proposition 9.19.
If $S>0$ [NB: this is just an abbreviation for positive definiteness] then $\ker S=\{0\}$. As a consequence, a positive definite linear system $Su=f$ with $f\in\operatorname{range}S$ must have a unique solution.
The first bit, $\ker S=\{0\}$ I understand. However, the next bit I don't get. I'm assuming $f\in\operatorname{range}S$ just means that the equation has at least one solution. But why the positive definiteness of $S$ implies this solution is unique? I don't know. The book has just finished covering the Fredholm Alternative, but since we have no information about the adjoint $S^*$ I don't think it helps.
At least in the finite dimensional case, if $\mathbf{S}$ is a positive definite matrix then it is definitely invertible, so the system $\mathbf{S}\boldsymbol{u}=\boldsymbol{f}$ definitely has a unique solution since we can just write $$\boldsymbol u=\mathbf{S}^{-1}\boldsymbol f$$ But I don't see how to generalize this to the infinite dimensional case.
Thanks for any help.
If $Su=f$ and $Sv=f$ then $S(u-v)=0$. Which is to say $u-v\in \ker S$.