For we define and . For which does the series
We may assume that is not the zero vector and otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.
(a) If the above condition is satisfied then, without loss of generality, let be the component of maximal absolute value and let . Hence , as
and the given series is convergent because .
(b) If the above condition is not satisfied, then there are at least components of maximal absolute value and therefore
and the given series is not convergent because .
Let be endowed with the usual product and the usual norm. If then we define . Prove that
Let . Prove that
Let . is twice differentiable with
Hence is convex. The tangent at has equation . The result follows.
The domain of is . Let us consider the logarithmic of
and differentiate it; Hence,
For we have
Thus, since .