A positive operator $P: H \to H$ is bounded below iff $P \ge cI$ for some $c>0$

Question

Let $H$ be a Hilbert Space. A positive operator $P: H \to H$ is bounded below iff $P \ge cI$ for some $c>0$.

I have been trying to prove this for a while but perhaps I am missing something:

First suppose that $P$ is bounded below, that is, there is some $c>0$ we have that $\lVert P x \rVert \ge c \lVert x \rVert$ for each $x\in H$. We need to show that $\langle Px-cx,x \rangle \ge 0$ for each $x\in H$. To this end, let $x\in H$. From $\lVert Px \rVert \ge c \lVert x \rVert$, by squaring both sides, we can conclude that $P^2-c^2I \ge 0$. I am expecting that $P \ge cI$ by taking square root but I cannot justify this argument. In fact, it is rather easy to see that $P: H \to H$ is bounded below iff $P^2 \ge c^2I$ for some $c>0$.

Hints on proving this will be appreciated!

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Here's one direction of the proof that I was able to check:

$(\Longleftarrow)$ Suppose that $P-cI \ge 0$. We claim that $\lVert Px \rVert \ge c \lVert x \rVert$ for each $x\in H$. If $x=0$, then the inequality is trivial, so, let $x\in H$ be nonzero.

\begin{align*} \lVert Px \rVert \rVert x\rVert &\ge \langle Px, x \rangle \ge \langle cx ,x\rangle \ge c \lVert x \rVert^2. \end{align*}

Cancelling $\lVert x \rVert$, we get what we wanted.

Answer 1

Any positive operator $P$ admits the square root, the positive operator $A$ so that $A^2=P.$ Hence $$\|Px\|^2=\|A(Ax)\|^2\|\le \|A\|^2\|Ax\|^2\\ =\|A\|^2 \langle Ax,Ax\rangle =\|A\|^2 \langle A^2x,x\rangle \\ =\|A\|^2\langle Px,x\rangle $$

Answer 2

Define a bilinear form $\psi:H\times H\to :\mathbb{R}, (x, y) \to \langle Px, y\rangle$, then you can verify that this new bilinear form satisfies C-S inequality : $$ \forall x, y \in H:\langle Px, y\rangle ^{2}\leq \langle Px,x\rangle \langle Py, y\rangle $$ In particular, for $y=Px$, we get: $\vert \vert Px \vert \vert^{4} \leq \langle Px,x\rangle \vert \vert Px \vert \vert^{2} \vert \vert P \vert \vert $ So if $P$ is bounded below, that is $\exists a>0,\vert \vert Px\vert \vert \geq a\vert \vert x\vert \vert $, then: $a^{2}\vert \vert x\vert \vert^{2}\leq \langle Px,x\rangle \vert \vert P \vert \vert $ So it suffice to take $c=\frac{a^{2}} {\vert \vert P\vert \vert } $