Banach Spaces, Bounded Linear Operator is Dot Product

Bounded Linear Operator is Dot Product

Let S be a separable hilbert space, with a complete orthonormal basis b₁ b₂ b₃ etc. Let f be a linear function from S into R. Assume f is continuous, or bounded. (Since S is a normed vector space, these are equivalent.)

Let c_i = f(b_i). Let v_n be the sum of c_ib_i, as i runs from 1 to n. Use the linear properties of f to show f(v_n) = the sum of c_i², as i runs from 1 to n.

Since f is bounded, let k be a bound on f. Now |f(v_n)| is no larger than k×|v_n|. Replace |v_n| with the square root of the sum of squares. Put this all together and get this.

f(v_n) ≤ k × sqrt(f(v_n))

f(v_n) ≤ k²

This applies in the limit, hence c_i forms a square summable sequence. Let u be the infinite sum of c_ib_i.

Let a_ib_i be the representation of an arbitrary element x in S. Since f is continuous, f(x) becomes the limit of f applied to the partial sums. This becomes the sum over a_ic_i. The same formula appears if you evaluate the dot product of x and u. Therefore f(S) = S.u.

Let x have norm 1, and apply cauchy schwarz, and x.u has norm bounded by |u|. The bound on f is no larger than |u|. Now f(u) = u.u = |u|², hence the bound on f is precisely |u|.

Suppose S.u, a continuous linear map, equals S.v for some other vector v. apply u-v to S and get 0. However, (u-v).(u-v) is nonzero, hence each linear function from S into R is S.u for a unique u. The bound on such a function is |u|, and is realized when u is dotted with itself.