I'm searching for an alternative proof of the following:
Let $U$ be a self-adjoint operator on a Hilbert space $H$, define $m=\inf_{\|x\|=1}\langle Ux,x\rangle$ and $M=\sup_{\|x\|=1}\langle Ux,x\rangle$. If $P$ is a
polynomial such that $P(x)\geq 0$ for $x\in[m,M]$, prove that $P(U)$
is a positive operator.
The usual proof is by using that $\sigma(P(U))=P(\sigma(U))$ (Spectral mapping theorem for polynomials). The intention is to prove this result without this theorem.
Edit: We say that $U$ is a positive operator if $\langle Ux, x\rangle\geq 0$ for every $x\in H$.
Best Answer
We use the following fact:
let $P\in\mathbb R[t]$ be positive on $[m,M]$. Then there exist polynomials $f_1,f_2,g_1,g_2,h_1,h_2\in\mathbb R[x]$ such that $$P(t)=(f_1(t))^2+(f_2(t))^2+(M-t)[(g_1(t))^2+(g_2(t))^2]+(t-m)[(h_1(t))^2+(h_2(t))^2].$$
From the definition of $m$ we get $\langle Ux,x\rangle\geq m\langle x,x\rangle$ for all $x,$ i.e. $U-m I$ is positive. Similarly, $M I-U$ is positive.
Hence for $x\in H$ we have $$\langle P(U)x,x\rangle=\langle (f_1(U))^2x,x\rangle+\langle (f_2(U))^2x,x\rangle+\dots \\ = \langle f_1(U)x,f_1(U)x\rangle+\langle f_2(U)x,f_2(U)x\rangle+\\ \langle (M-U)g_1(U)x,g_1(U)x\rangle+\langle (M-U)g_2(U)x,g_2(U)x\rangle+\\ \langle (U-m)h_1(U)x,h_1(U)x\rangle+\langle (U-m)h_2(U)x,h_2(U)x\rangle\geq 0$$