2

I have been assigned to do the question I've attached. I have managed to do a,b, and c. Now I have 2 questions: (I'll use normal brackets for inner product brackets)

Firstly, in part (a), I used that lim as n-->∞ of (An(x)-A(x)|y) = ∞ of: (An(x)-A(x)|y), and I don't know how to prove this continuity property of the inner product.

I have been having difficulties with (d): I can show that any element of w is in N⊥, since for any m in N and w in W, |(m|w)|=|(An(m)|An(w))| (since U preserves length, and consequently so does An), and this tends to (0,w)=0 as n tends to infinity, using results in b and c (again I don't know how to prove continuity). My problem is, how to show that any element not in W, is not in N⊥. How do I proceed? Thanks! edit: picture of question attached

lkjhgfdsa
  • 591

1 Answers1

1

Note that $V=N\oplus N^{\perp}= \text{Im }(I-U)\ \oplus \ker{(I-U)}$. Since $N=\text{Im }(I-U)$ by definition, then it must be the case that $N^{\perp}=\ker{(I-U)}$. Hence, if $v$ is a vector in $N^{\perp}$, then $(I-U)v=0$, which means that $v\in W$.

As for the continuity of $\langle \_,\_ \rangle$, use the fact that $f(y)=\langle y,x \rangle$ for fixed $x$ defines a linear functional, which is bounded by the Cauchy-Schwarz inequality. Finally, one can prove that a bounded linear functional is continuous. The details of these two stages can be found in the links below:

Is Inner product continuous when one arg is fixed?

https://en.wikipedia.org/wiki/Bounded_operator#Equivalence_of_boundedness_and_continuity

Exp5LogMingus
  • 152