In the second book on Operator Algebras and Quantum Statistical Mechanics by Bratteli and Robinson (on page 11) they are looking at the algebra of fermionic creation and annihilation operators generated by some Hilbert space. These operators act on the fermionic Fock space, and it is clear to me why they are bounded. (Actually they are defined on a dense subset, but boundedness implies the existence of unique bounded extension).
However they use an argument which I do not think works directly, the CAR relations are:
$$\{a(f),a^*(f)\}=||f||^2 \qquad \qquad \{a(f),a(f)\}=0=\{a^*(f),a^*(f)\}$$
Where $f$ is a state in the original Hilbertspace. So:
$$(a^*(f)a(f))^2=a^*(f)a(f)a^*(f)a(f)=a^*(f)\{a(f),a^*(f)\}a(f)=||f||^2a^*(f)a(f)$$
Where the anti-commutator can be inserted since $a^*(f)^2=0$. Now they use compatibility of the norm with the * operation:
$$||a(f)||^4=||(a^*(f)a(f))^2||=||f||^2||a^*(f)a(f)||=||f||^2||a(f)||^2$$
And we would get result $||a(f)||=||f||$ and $a(f)$ is bounded. However we must assume explicitly that $a(f)$ is bounded, i.e. $||a(f)||$ exists, in order to use the compatibility with the involution. So in my eyes the proof, as I have written it here, is false.
Is there another way the second line implies boundedness?
One final comment, which may actually be essential, in the book the wording they use is that this proves that the operator $a(f)$ has a bounded extension. But if an operator that is defined on a dense subspace is bounded it automatically has a bounded extension, and the other way around also, so I am taking these two statements to be equivalent.