If then . For convenience, suppose that the coordinates of x are all positive, so that we can drop the absolute value symbols in the formula . Then , where . But . Thus and . Now let . Then and

That shows that (which is a vector with the same components as x, but with the norm given as a functional on ) has norm at least (which is what you wanted).

It remains to deal with the fact that the components of x may not all be positive. If the scalars are real then you just have to ensure that, for each j, has the same sign as . (Then all the terms in the sum will be positive.) If the scalars are complex then you have to multiply each by an appropriate complex number of absolute value 1 to ensure the same result.