Proof if $T \in L \left( V,V \right)$ is self-adjoint and b,c $\in \mathbb{R}$ such that $b^2 < 4c$ , then $T^2 + bT + cI$ is a positive operator.

Question

I also have a hint: Use Cauchy–Schwarz inequality.
I'm really stuck with this problem. I think that $T^2 + bT + cI \geq 0$ because of $b^2 - 4c < 0$ may help but I don't know how to relate it with the hint.

Answer 1

Here's an approach without the Cauchy–Bunyakovsky–Schwarz inequality.

Because $b^2 < 4c$, the expression $x^2 + bx + c$ can be rewritten as $(x-h)^2 + k$ for some $h \in \Bbb R$ and $k > 0$ (via "completing the square"). Equivalently, we have $$ T^2 + bT + cI = (T - h I)^2 + k I. $$ Because $T$ is self-adjoint, $(T - hI)^2$ is a non-negative operator. Moreover, $k I$ is a positive operator. The sum of a non-negative operator and a positive operator is necessarily positive.

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An approach using CBS: for $x \in V$ with $x \neq 0$, we have $$ \langle x, (T^2 + bT + c I)x \rangle = \\ \langle Tx,Tx\rangle + b \langle x, Tx\rangle + c\langle x,x \rangle =\\ \|Tx\|^2 + c \|x\|^2 + b\langle x,Tx\rangle \geq\\ \|Tx\|^2 + c \|x\|^2 - |b|\cdot \|x\|\cdot \|Tx\| = \\ \|x\|^2 \left[ \left(\frac{\|Tx\|}{\|x\|} \right)^2 - |b| \cdot \left(\frac{\|Tx\|}{\|x\|} \right) + c \right]. $$ From the fact that $|b|^2 - 4c < 0$, conclude that the above expression is positive.

Answer 2

$\begin{align}\langle (T^2+bT+cI)v,v\rangle&=\langle T^2v,v\rangle+b\langle Tv, v\rangle +c\langle v, v\rangle\\&=\langle Tv, Tv\rangle +b\langle Tv, v\rangle+c\langle v, v\rangle \\&\ge \|Tv\|^2-|b|\|Tv\|\|v\|+c\|v\|^2\\&=(\|Tv\|-\frac{|b|\|v\|}{2})^2+(c-\frac{b^2}{4})\|v\|^2\\&\ge 0 \space (\text{in fact} >0 ) \end{align}$

1. Since $T^{\star}=T$

$\langle T^2v, v\rangle =\langle Tv, T^{\star}v\rangle=\langle Tv, Tv\rangle$

2. $|b \langle Tv, v\rangle |\le |b| \|Tv\|\|v\|$ ( Cauchy-Schwarz)

Implies $b \langle Tv, v\rangle\ge -|b| \|Tv\|\|v\|$

3. $ b^2-4c<0$ implies $c-\frac{b^2}{4}>0$

Answer 3

With spectral theory: Let $V$ be a Hilbert space. For a bounded linear operator $A$ on $V$ let us denote the spctrum of $A$ by $ \sigma (A).$

Let $p$ the plynomial $p(z)=z^2+bz+c.$ Since $b^2 - 4c < 0$, we have that $p(x) >0$ for all $x \in \mathbb R.$

Now let $z \in \sigma(T^2+bT+cI).$ The spectral mapping theorem shows that $z=p(x)$ for some $x \in \sigma(T).$ Since $T$ is self-adjoint, $x \in \mathbb R.$ Hence $z>0.$

This shows that $T^2+bT+cI$ is positive.