add, after (*), then we can finish the proof. Conclusion: the subsequence of f_n which converges in every point of the sequence x_j is a Cauchy sequence in C([0,1]). This completes the proof. Right blackboard above refers to http://www.few.vu.nl/~jhulshof/FA2008/week3/ Above: f_n is defined by f_n(x)=x^n. With respect to |.| this sequence has no limit in C([0,1]). The obvious limit is the zero function, which differs from the pointwise limit only in x=1. In L^2(0,1) we do have that the limit is the zero function because in that space f=g means, by definition, that f(x)=g(x) for almost every x in [0,1]. Thursday:

Missing blackboard, on which I explained that the closure of W_1 contains an open ball with center in the origin. See homework 3. Denoting the greek x by xi, for every xi_0 sufficiently small as specified above, we can find xi_1 with the 3 properties specified above,

NO MATTER what we choose for lambda.

Below we apply this repeatedly, to construct a sequence x_n, such that x_1+...+x_n approaches x_0 AND such that A(x_1+...+x_n)=Ax_1+...+Ax_n also converges, with limit (SINCE A IS CLOSED) equal to Ax_0. Now read carefully the last blurred sentence in the corner below on the right. It gives a bound on ||Ax_0|| that follows from the geometrical series. See also the exercises in homework 3 related to steps in this proof. The only thing we used on x_0 was that ||x_0|| is smaller than r_0. Thus, with this bound, A is bounded on a ball and we are DONE.