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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!

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  • $\begingroup$ 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? $\endgroup$ – Keith McClary Jan 28 '19 at 17:43
  • $\begingroup$ @KeithMcClary Nonzero and positive doesn't make them strictly positive. $\endgroup$ – Aweygan Jan 28 '19 at 18:42
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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.

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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 $\langle u,Bu\rangle>0$, $$\langle u+v,B(u+v)\rangle=\langle u,Bu\rangle>0$$

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