If $A$ is a nonnegative self-adjoint operator on a Hilbert space $H$, can we show that $\left\|Ax\right\|_H^2=\langle Ax,x\rangle_H$?

Question

Let $H$ be a $\mathbb R$-Hilbert space and $A\in\mathfrak L(H)$ be nonnegative and self-adjoint and $x\in H$.

Are we able to show that $\left\|Ax\right\|_H^2=\langle Ax,x\rangle_H$?

Clearly, $$\langle Ax,x\rangle_H=\left\|A^{\frac12}x\right\|_H^2.\tag1$$ On the other hand, for each self-adjoint $B\in\mathfrak L(H)$, $$\left\|B^2\right\|_{\mathfrak L(H)}=\left\|B\right\|_{\mathfrak L(H)}^2\tag2.$$ However, the claim is not immediate from $(2)$.

Answer 1

The only bounded operator such that and $||Ax||_H=\langle Ax, x \rangle_H$ for all $x$ is the zero operator. Proof: replace $x$ by $n$ and divide by $n^{2}$. You get $\frac 1 n ||Ax||_H=\langle Ax, x \rangle_H$. Let $n \to \infty$ to get $\langle Ax, x \rangle_H=0$. Go back to the given equation to get $Ax=0$. This is true for every $x$ so $A=0$.

Answer for the edited version. Replace $A$ by $nA$, divide by $n^{2}$ and let $n \to \infty$.