| dbp:proof
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- Suppose that is bounded. Then, for all vectors with nonzero we have
Letting go to zero shows that is continuous at
Moreover, since the constant does not depend on this shows that in fact is uniformly continuous, and even Lipschitz continuous.
Conversely, it follows from the continuity at the zero vector that there exists a such that for all vectors with
Thus, for all non-zero one has
This proves that is bounded. Q.E.D. (en)
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