Proof of Cauchy-Schwarz Inequality for a Positive Semi-Definite Self-Adjoint Operator

Let $u,v\in V$ and write $a=\langle u,T(u)\rangle$, $b=\langle v,T(v)\rangle$ and $z=\langle u,T(v)\rangle$. By Positive Semi-Definite Operator, $a$ and $b$ are real numbers with $0\le a$ and $0\le b$. Let $\overline{w}$ denote the complex conjugate of a complex number $w$.

We first record three facts.

(i) $\langle v,T(u)\rangle=\overline{z}$. Indeed, $T$ being self-adjoint gives $\langle T(v),u\rangle=\langle v,T(u)\rangle$, while conjugate symmetry (condition 1 of Complex Inner Product Space) gives $\langle T(v),u\rangle=\overline{\langle u,T(v)\rangle}=\overline{z}$.

(ii) For all $x,y,y'\in V$ and every complex $t$, $\langle y+ty',x\rangle=\langle y,x\rangle+\overline{t}\,\langle y',x\rangle$. This follows by applying conjugate symmetry, then conditions 2 and 3 of Complex Inner Product Space in the second argument, then conjugate symmetry again, using that conjugation preserves sums and products.

(iii) For every complex $w$, $w\overline{w}=|w|^{2}$, where $|w|$ is the modulus. Indeed, writing $w=p+qi$ with $p,q$ real, claim 4 of Canonical Form and Arithmetic of Complex Numbers gives $w\overline{w}=p^{2}+q^{2}$, which is $|w|^{2}$ by the definition of the modulus.

Claim 1. Let $t$ be a complex number. Since $T$ is a linear operator, $T(u+tv)=T(u)+tT(v)$. Using (ii) in the first argument and conditions 2 and 3 of Complex Inner Product Space in the second, together with (i),

By Positive Semi-Definite Operator the left-hand side is a real number that is at least $0$.

If $z=0$, then $|z|^{2}=0$ and $0\le ab$ because $a$ and $b$ are nonnegative, so the asserted inequality holds. Assume therefore $z\ne0$; then $|z|\ne0$, and $0<|z|$ since a modulus is nonnegative. Let $s$ be a real number and put $t=-(s/|z|)\overline{z}$. By (iii),

and $\overline{t}\,\overline{z}=\overline{tz}=-s|z|$ because $-s|z|$ is real. Substituting into the display,

If $b=0$, then taking $s=(a+1)/(2|z|)$ gives $2s|z|=a+1$ and hence $0\le a-(a+1)=-1$, which is false in the ordered field $\mathbb{R}$. Hence $b\ne0$ and so $0<b$. Taking $s=|z|/b$ gives

and multiplying by the positive number $b$ yields $0\le ab-|z|^{2}$, that is $|z|^{2}\le ab$, which is claim 1.

Claim 2. Suppose $\langle u,T(u)\rangle=0$, and let $v\in V$ be arbitrary. Claim 1 gives

Writing $\langle u,T(v)\rangle=p+qi$ with $p,q$ the real and imaginary parts, the modulus satisfies $|p+qi|^{2}=p^{2}+q^{2}$, and squares are nonnegative in an ordered field, so $p^{2}+q^{2}\le0$ forces $p^{2}=q^{2}=0$ and hence $p=q=0$. Thus $\langle u,T(v)\rangle=0$ for every $v\in V$.

Since $T$ is self-adjoint, $\langle T(u),v\rangle=\langle u,T(v)\rangle=0$ for every $v\in V$. Taking $v=T(u)$ gives $\langle T(u),T(u)\rangle=0$, so $T(u)=0_{V}$ by condition 4 of Complex Inner Product Space.