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Let $H$ be a Hilbert space and $T\in \mathcal B(H)$ be a bounded, self-adjoint linear operator that is positive in the sense that $\sigma(T) \subset [0,\infty)$.

Is there an elementary method of proving that $T$ induces a positive semidefinite quadratic form, i.e. $$ \langle Tx,x\rangle \ge 0 $$ for all $x\in H$?

The proof of this statement (and its converse) can be found in this post. However, while the converse can be proved by an elementary mean, the proof of the statement that I want relies on the spectral theorem for self-adjoint operators. I want to know if there is a more rudimentary way to do it (i.e. without using these high-tech theorems).

BigbearZzz
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1 Answers1

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Theorem: Let $T$ be a bounded self-adjoint operator on a Hilbert Space $H$. Then $\lambda=\inf_{\|x\|=1}\langle Tx,x\rangle$ is in the spectrum of $T$. Therefore, if $\sigma(T)\subseteq[0,\infty)$, then $\langle Tx,x\rangle \ge 0$ for all $x$.

Proof: Suppose $T$ is bounded and self-adjoint with $\lambda = \inf_{\|x\|=1}\langle Tx,x\rangle$. Then $\langle (T-\lambda I)x,x\rangle \ge 0$, which is enough to imply that the Cauchy-Schwarz inequality holds for the form $[x,y] = \langle (T-\lambda I)x,y\rangle$. Writing this out, $$ |[x,y]|^2 \le [x,x][y,y] \\ |\langle (T-\lambda I)x,y\rangle|^2 \le \langle(T-\lambda I)x,x\rangle\langle (T-\lambda I)y,y\rangle $$ Now let $y=(T-\lambda I)x$: $$ \|(T-\lambda I)x\|^4 \le \langle (T-\lambda I)x,x\rangle\|T-\lambda I\|\|(T-\lambda I)x\|^2 \\ \|(T-\lambda I)x\|^2 \le \|T-\lambda I\|\langle (T-\lambda I)x,x\rangle $$ Let $\{x_n\}$ be a sequence of unit vectors such that $\lim_n\langle Tx_n,x_n\rangle=\lambda$. It follows that $\lim_n \|(T-\lambda I)x_n\|=0$, which puts $\lambda\in\sigma(T)$. $\;\;\blacksquare$

Disintegrating By Parts
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