# Strictly positive inner product for a pair of non-zero, positive operators.

Let $$A,B$$ be non-zero positive operators on a infinite-dimensional separable Hilbert space $$(H , \langle \cdot, \cdot \rangle)$$. I am required to prove that there exists $$u' \in H$$ such that \begin{alignat*}{2} \langle Au' , u'\rangle >0 \ \ \text{and} \ \ \langle Bu', u' \rangle >0. \end{alignat*} I am quiet stuck with this problem. For a few things I have tried. It is obvious that there exists $$v,w \in H$$ such that \begin{alignat*}{2} \langle Av, v \rangle > 0 \ \ \text{and} \ \ \langle Bw, w\rangle>0. \end{alignat*} And I have tried to calculate \begin{alignat*}{2} \langle A(v + w), v +w \rangle \end{alignat*} for which I have to now show for instance $$\text{Re} \langle Av, w \rangle \geq 0$$. Alternatively I could try to directly find a positive operator $$E$$ such that \begin{alignat*}{2} \langle Eu \ , \ u\rangle \leq \langle Au, u \rangle \ \ \text{and} \ \ \langle Eu , u \rangle \leq \langle Bu, u \rangle. \end{alignat*} I have also tried to apply orthogonal projections and polarisations, etc, but to no success. Hopefully it's some trivial details which I have missed.

The spectral theorem for bounded self-adjoint operator is not at my disposal.

Could anyone provide me with some hint? Thanks!

• I must be missing something here. If these are strictly positive bounded (so defined on the whole space) operators, doesn't every vector satisfy that? – Keith McClary Jan 28 '19 at 17:43
• @KeithMcClary Nonzero and positive doesn't make them strictly positive. – Aweygan Jan 28 '19 at 18:42

Let $$v,w\in H$$ be as you have defined them. For $$t\in[0,1]$$ put $$x_t=tu+(1-t)w$$, and define $$f,g:[0,1]\to [0,\infty)$$ by $$f(t)=\langle Ax_t,x_t\rangle,\quad g(t)=\langle Bx_t,x_t\rangle.$$
Note that $$f$$ and $$g$$ are non-zero polynomials (of degree at most $$2$$). Argue that there is some point $$t_0\in[0,1]$$ such that both $$f(t_0)>0$$ and $$g(t_0)>0$$, which proves the result.
Suppose $$\langle u,Au\rangle=0$$ and $$\langle v,Av\rangle>0$$. Then from $$0\le\langle u+tv,A(u+tv)\rangle=2t\mathrm{Re}\langle u,Av\rangle+t^2\langle v,Av\rangle$$ it follows that $$\mathrm{Re}\langle u,Av\rangle=0$$. Similarly, by replacing $$t$$ by $$it$$, we get $$\mathrm{Im}\langle u,Av\rangle=0$$, so $$\langle u,Av\rangle=0$$ whatever $$v$$. Hence $$\langle u+v,A(u+v)\rangle=\langle v,Av\rangle>0$$
Similarly for $$B$$, assuming $$\langle v,Bv\rangle=0$$ and $$\langle u,Bu\rangle>0$$, $$\langle u+v,B(u+v)\rangle=\langle u,Bu\rangle>0$$