Let be an inner product space, be a linear operator such that show that is injective.
The obvious reason is that to go notice we were reducing a bivalued function to a single valued function. There is an obvious way to do this by replacing both arguments of the bivalued function by one value. To go the other way is not so easy.
So, my point is that in general and we only need the latter opposed to the former, so it's strange they gave us an "extra" condition.
For fun, see if you can find an easy inner product using the norm with
EDIT: An easy way to think about this is and . One clearly follows from the other but not vice versa.