As mentioned also in Bounded and Unbounded Operator, an operator $A$ is said to be bounded, if $$\|Af\|\leq k \|f\|,$$ where the constant $k$ does not depend on the choice of $f$ (let us consider a map to the same Banach space).

However, in a mathematical physics text I came across a definition: a symmetric operator $B$ is said to be bounded from below if there $\exists$ a constant $c$ such that $$\langle\psi,B\psi\rangle\geq c\|\psi\|^2$$ for all $\psi$ in the domain of $B$.

Both definitions are logical (in the second one we can imagine $B$ being the Hamiltonian, than the system energy is bounded from below and hence the system is stable).

The only think that bothers me is when we rewrite the first definition into a similar form to the second one (we assume the norm comes from an inner product), namely:

$$\langle Af, Af\rangle \leq k\|f\|^2,$$ we get something quite different on the left-hand side, so the same words (bounded operator) refer to different things. Any hints how I can clarify this to myself?


3 Answers 3


TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia.

In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$. It follows that $\langle A \rangle_{\psi}\in\mathbb{R}$ is real.

  1. That $A$ is bounded from below means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\geq ~C. $$

  2. That $A$ is bounded from above means that $$\exists C\in \mathbb{R}~ \forall \psi\in D\backslash\{0\}: ~~ \langle A \rangle_{\psi}~\leq ~C. $$

  3. That $A$ is bounded means that $$\exists C\geq 0~ \forall \psi\in D\backslash\{0\}: ~~ \frac{||A \psi||}{||\psi||}~\leq ~C, $$ which is equivalent to $A^{\dagger}A$ $(=A^2)$ being bounded from above, which in turn is equivalent to $A$ being bounded from both above and below.

  • $\begingroup$ also relative boundedness: planetmath.org/katorellichtheorem $\endgroup$
    – Phoenix87
    Feb 12, 2015 at 13:49
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Feb 12, 2015 at 22:39
  • $\begingroup$ Could you list some examples of bounded operators used in quantum mechanics? $\endgroup$
    – Seeker
    May 20, 2017 at 8:00

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!

  • $\begingroup$ +1 Thanks! Your answer is very useful, also by providing context and relation to other stuff which I need to think through. $\endgroup$
    – wondering
    Apr 24, 2015 at 21:38

When one says that an operator is bounded, you can think of it as being bounded from above. This is different from being bounded from below. An operator can be bounded (from above) and bounded from below, or perhaps just bounded, or just bounded from below.

Observe that $(Af,Af)$ and $(\psi,B\psi)$ are slightly different: the former is always positive for any operator $A$, not necessarily symmetric, while the latter can be negative. If $c$ turns out to be non-negative then the operator $B$ is positive and, if it is also bounded (from above) its spectrum is contained in $[c,\Vert B\Vert]$.

  • $\begingroup$ Can you tell me why $\hat B$ is required to be symmetric? And, symmetric in which sense, i.e. with respect to what? Should it be equal to its transposed? Should it be also real ? $\endgroup$
    – Sofia
    Feb 12, 2015 at 13:35
  • $\begingroup$ symmetric here is a synonym of Hermitian, i.e. an operator that looks like a self-adjoint operator, but it is not defined everywhere (but it is usually densely defined) $\endgroup$
    – Phoenix87
    Feb 12, 2015 at 13:38
  • $\begingroup$ I understand. Your explanation is very clear. $\endgroup$
    – Sofia
    Feb 12, 2015 at 13:44
  • $\begingroup$ Thank you, very helpful as well! I cannot upvote +1 as yet, sorry, I'll do it later! $\endgroup$
    – wondering
    Feb 14, 2015 at 1:02

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