Continuity(?) of a linear map $\alpha: H \to B(H)$ satisfying canonical anti-commutation relations Let $H$ be a separable Hilbert space and $\alpha: H \to B(H)$ a linear map s.t. $\forall f, g \in H$:

*

*$\alpha(f)\alpha(g) + \alpha(g)\alpha(f) = 0$

*$\alpha(f)^*\alpha(g) + \alpha(g)\alpha(f)^* = \langle g,f \rangle I$
If $\alpha$ is continuous, we get that $C^*(\alpha(f))$ is the CAR algebra. But does continuity follow automatically from the above conditions? Or do we need to specify it in the assumption(s) to generate the CAR algebra?
It seems to me that continuity should probably follow from the second condition, but I've been playing around with it and can't quite make it work.

The second condition applied to $f=g$ is
$$
\alpha(f)^*\alpha(f) + \alpha(f)\alpha(f) ^* = \|f\|_H^2 \cdot Id.
$$
Let $x\in H$. Then
$$
\|\alpha(f)x\|_H^2 +\|\alpha(f)^*x\|_H^2 
= \langle x, (\alpha(f)^*\alpha(f) + \alpha(f)\alpha(f) ^* )x \rangle
= \|f\|_H^2 \|x\|_H^2,
$$
where the first equality uses the definition of the norm, and the second one uses the first equation above.
This proves $\|\alpha (f)\|_{B(H)} \le \|f\|_H$, and $\alpha$ is bounded, hence continuous.
