Actually, the conditions $$\langle x|A x \rangle \geq a \langle x |x \rangle$$ and $$\langle x|A x \rangle \leq b \langle x |x \rangle$$ with $a,b \in \mathbb R$ fixed and all $x\in D(A)$ (the domain of $A$) refer to **boundedness **(resp. from below or from above) of the** quadratic form **associated to the linear operator $A$. This operator can always be supposed to be Hermitian since the anti-Hermitian part does not play any role in $\langle x|A x \rangle$.

However, making stronger the hypotheses and assuming that $D(A)\subset {\cal H}$ is a dense linear manifold in the Hilbert space ${\cal H}$ and that $A=A^*$, namely, $A$ is self-adjoint, then the (both from below and from above) boundedness condition of the quadratic form associated to $A$ is **equivalent** to that of the operator $A$ itself.

Indeed, if the operator $A$ is bounded, it must be $D(A)={\cal H}$ (well-known property of self-adjoint operators) and

$$| \langle x |A x \rangle | \leq ||x|| ||Ax|| \leq ||x|| ||A|| ||x|| = ||A|| ||x||^2$$

so that the quadratic form is bounded both from below and from above with $a = -||A||$ and $b=||A||$.

If, conversely the quadratic form is bounded, the spectral theorem immediately implies that

$$\lambda - a \geq 0\quad\mbox{and}\quad b- \lambda \geq 0\quad \mbox{if $\lambda \in \sigma(A)$,}$$

so that the spectrum $\sigma(A)$ is included in the bounded interval $[a,b]$. The spectral radius formula, in turn, implies that

$$||A|| \leq \max\{|a|, |b|\} < + \infty\:.$$

The two notions of boundedness, for operators representing **observables** in QM, are therefore equivalent. This is particularly remarkable because the boundedness of the quadratic form of $A$ is equivalent to the boundedness of the spectrum (see above) of $A$, which means that the values attained by the observable $A$ form a bounded set. This equivalence shows in particular that most observables (think of position or momentum) cannot be represented by bounded operators (since the attained values are arbitrarily large) and this is one of the reasons why QM is technically complicated mathematically speaking, the reason being physical however!