Infinite-dimensional inner product space: if $A \geq 0$ and if $\langle Ax, x\rangle = 0$ for some $x$, then $Ax = 0$.

Exercise 8, Section 82 from PR Halmos's Finite-Dimensional Vector Spaces, 2nd Edition

If $$A$$ is a positive semidefinite operator, and if $$\langle Ax, x\rangle = 0$$ for some vector $$x$$, show that $$Ax = 0$$. The underlying inner product space is not specified as finite-dimensional. The scalar field is not specified as real or complex.

I am able to establish the assertion assuming that the inner product space is finite-dimensional. Struggling with extending the argument to infinite-dimensional spaces however.

My argument for the finite-dimensional case goes as follows. Section 82 ("Functions of Transformations") of the book argues that every positive operator on a finite-dimensional inner product has a positive square root (function) associated. Thus, we observe that $$0 = \langle Ax, x\rangle$$ $$= \langle \sqrt A \sqrt Ax, x\rangle$$ $$= \langle \sqrt Ax, {\sqrt A}^*x\rangle$$ $$= \langle \sqrt Ax, \sqrt Ax\rangle$$ $$= \Vert \sqrt Ax \Vert^2$$ $$\implies$$ $$\sqrt Ax = 0$$ $$\implies \sqrt A \sqrt Ax = 0$$ $$\implies Ax = 0$$.

Unclear on how to extend this argument to the infinite-dimensional case. Would appreciate an advice. Thanks.

• Does Halmos' definition of positive (semi)definite include the assumption that $A$ is self-adjoint ($A^* = A$)? Some authors assume this and others do not.
– user169852
Jun 23, 2020 at 1:59
• Yes. Positive semidefinite and positive definite operators are self-adjoint by definition, accordingly to Halmos. Jun 23, 2020 at 2:02
• Every positive self-adjoint operator has a square root, no matter if the underlying Hilbert space is finite-dimensional or not (if the space is not complete, it might map to the completion, but that does not matter for your proof). But I guess you don't know that yet? Jun 23, 2020 at 5:51
• The question doesn't say that the underlying inner product space is a Hilbert space (i.e., a complete inner product space). The book introduces the concept of Hilbert spaces much later. So, like elsewhere in the book, the space is to be understood as a not-necessarily-finite-dimensional space. Jun 23, 2020 at 6:37
• @Omnomnomnom Why? It certainly has a square root in the completion, but why should it map the inner product space into itself? Jun 23, 2020 at 7:52

The map $$(y,z)\mapsto \langle Ay,z\rangle$$ is a semi-inner product (i.e. it satisfies the same conditions as an inner product except for positive definiteness, which is replaced by positive semi-definiteness). In particular, the Cauchy-Schwarz inequality applied to $$y=x$$ and $$z=Ax$$ gives: $$\lvert \langle Ax,Ax\rangle\rvert\leq \langle Ax,x\rangle^{1/2}\langle A(Ax),Ax\rangle^{1/2}=0.$$ Thus $$Ax=0$$.
Suppose that $$(Ax,x) = 0$$, but $$Ax \neq 0$$. Consider the vector $$v = Ax + tx$$, with $$t \in \Bbb R$$. We have $$(Av,v) = (A^2x+tAx,Ax+tx) = (x,Ax)t^2 + [(A^2x,x) + (Ax,Ax)]t + (A^2x,Ax)\\ = 2\|Ax\|^2 t + (A^2x,Ax).$$ We see that for a "sufficiently negative" $$t$$ ($$t < -\frac{(A^2x,Ax)}{2\|Ax\|^2}$$), $$(Av,v)$$ must be negative. So, $$A$$ cannot be positive semidefinite.