Banach Spaces, Dot Product is Continuous

Cauchy Schwarz

The dot product is tied to the norm, and the norm satisfies certain properties, such as the triangular inequality. This can be used to prove cauchy schwarz in an arbitrary hilbert space. Use the properties of norm and dot product to write the following.

0 ≤ |x-ly|² =
(x-ly).(x-ly) =
x.x - 2l(x.y) + l²(y.y)

The inequality becomes equality iff x = ly.

Setting l = 0 or y = 0 gives 0 ≤ x.x, which isn't very interesting, so assume l > 0 and y ≠ 0, and write the following inequality.

2x.y ≤ x.x/l + ly.y

Set l = |x|/|y| and find x.y ≤ |x|×|y|. If x or y is 0 we have equality, and if x = ly we began with an equation, which produces equality. This is cauchy schwarz. The dot product is bounded by the product of the norms, with equality iff one vector is a linear multiple of the other. Remember that 0 is technically a linear multiple of x, and sure enough, 0.x = |0| times |x|.

Dot Product is Continuous

Fix a vector v and consider the linear map from v.S into R. Concentrate on the unit sphere in S, the vectors in S with norm 1. Now v.x is bounded by |v|×|x|. Thus v.S is a bounded operator, hence continuous.

Let T be S cross S, with the product topology. This is another banach space. The dot product now maps T into R. Select x and y from S such that the ordered pair x,y in T has norm 1. Thus |x| and |y| have norm at most 1 in S. The dot product x.y is bounded by 1. Once again a linear operator on a banach space is bounded, hence continuous. The dot product is a continuous map from S cross S into the reals.