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Let $H$ be an Hilbert space, and let $T$ be a self-adjoint, positive (and therefore bounded) operator $H \to H$, with $||T||<2012$. Let $P$ be a polynomial with real coefficients such that the minimum of $P$ on $[-2012,2012]$ is positive. Consider the operator $S=P(T)$. It is also self-adjoint and bounded. I ask : must it necessarily be positive (i.e. do we have $(Sx,x) \geq 0$ for every $x$)?

If $T$ is diagonalizable by an orthogonal basis $\cal B$, then $S$ will also be diagonal relatively to $\cal B$, and the eigenvalues of $S$ will be the numbers $P(\lambda)$, where $\lambda$ is an eigenvalue of $T$. So the answer is yes in this case.

This covers the case when $H$ is finite-dimensional. When $H$ is infinite-dimensional, however, things are not so clear (at least to me).

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You can argue in the infinite dimensional case along the same lines. Note first that a self-adjoint $S$ is positive exactly iff $\sigma(S) \subseteq [0,\infty)$. We have by the spectral theorem that $\sigma(S) = p\bigl(\sigma(T)\bigr)$. As $\sigma(T) \subseteq [-\|T\|, \|T\|]$, we have $\sigma(S) \subseteq[0,\infty)$ as $p$ is postive there.

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    Be careful with the equivalence positive iff spectrum positive - maybe you would better specify what you understand along the lines of positive...2014-08-28
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Maybe Dunford&Schwarz will help you.The order will be preserved when doing operator calculus to a self-adjoint operator.I'm not quite convinced,just check the book.