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Problem: Let $X$ be a vector space over $\mathbb{R}$ and $Y \subset X$ a linear subspace. Let $p: X \to \mathbb{R}$ be a sublinear functional and $f: Y \to \mathbb{R}$ linear with $f \leq p$ on $Y$ (by Hahn-Banach this can be extended of course)

Consider now $G \subset \mathcal{L}(X)$ a subset of bounded linear operators $T: X \to X$ with the properties that id$_X \in G$ and for all $A,B \in G, AB \in G$ and moreover $AB=BA$.
Assume that for all $A \in G$ we have $p(Ax) \leq p(x)$ for all $x \in X, Ay \in Y$ and $f(Ay)=f(y)$ for all $y \in Y$

Claim: There exists $F: X \to \mathbb{R}$ linear with $F_{ \mid Y} =f, F \leq p$ on $X$ and $F(Ax)=F(x)$ for all $x \in X$ and $A \in G$

My approach: I am given the hint to consider $q(x):= \inf_{A_1, \dots , A_n} \frac{1}{n}p(A_1x + \dots + A_n x)$ where the infimum is taken over finitely many $A_i \in G$, then show that $q$ is sublinear and $f(y) \leq q(y)$ for all $y \in Y$ in order to apply Hahn-Banach.

Showing that $q$ is sublinear follows from the fact that all the $A_i$ are linear and $p$ is sublinear by definition. Further for $y \in Y$ and $A_1, \dots , A_n \in G$ we have \begin{align} f(A_1y + \dots + A_ny) &= f(A_1y) + \dots + f(A_ny) =n f(y) \end{align} But also $f(A_1y + \dots + A_ny) \leq p(A_1y + \dots + A_ny)$, thus we have $f(y) \leq \frac{1}{n}p(A_1y + \dots + A_ny)$ taking the infimum we get that $f(y) \leq q(y)$ on $Y$.

By Hahn-Banach there exists $F: X \to \mathbb{R}$ linear with $F_{ \mid Y} = f$ and $F(x) \leq q(x)$ for all $x \in X$. However since $p(A_1x + \dots + A_nx) \leq p(A_1x)+ \dots + p(A_nx) \leq n p(x)$ it also follows that $q \leq p$ on $X$ which takes care of the first claim.

I need to show that $F(Ax)=F(x)$ for all $x \in X$ and $A \in G$ holds. It would be good if I could show that $F(Ax) \leq F(x)$, then the other inequality would follow by linearty of $F$. \begin{align}F(Ax) \leq q(Ax)&= \inf_{A_1, \dots , A_n} \frac{1}{n} p(A_1Ax + \dots + A_nAx) \\ &= \inf_{A_1, \dots , A_n} \frac{1}{n} p(AA_1x + \dots + AA_nx) \end{align} Here I get stuck. My Question(s) would be if my approach above is correct so far and how I would go on about concluding that $F(Ax)=F(x)$ on $X$.

Community
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Spaced
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1 Answers1

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Your approach is correct. Use a telescoping sum to show that $q(Ax-x)=0.$

Vogel
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  • I tried and I came to $$q(Ax-x)= \inf (1/n) p(A_1(Ax-x) + \dots + A_n(Ax-x)) = \inf (1/n) p(A_1 Ax - A_1x + \dots + A_nAx - A_nx)) = \inf (1/n) p(A A_1x - A_1 x + \dots + A A_n x - A_nx) = \inf (1/n) p (A(A_1x + \dots + A_nx) - (A_1x + \dots + A_nx)) $$ But I doubt that's what you wanted me to do. – Spaced Nov 05 '17 at 19:07
  • @Spaced: Try plugging in $A_i=A^{i-1}$ – Vogel Nov 05 '17 at 19:24
  • Thanks, I will obtain that $q(Ax-x)= \inf (1/n) p(A^n x -x)$ but the infimum is now taken over $id_x, A, A^2, \dots , A^{n-1}$ so I can drop it, the LHS side does not depend on $n$ and for the RHS pass to the limit? – Spaced Nov 05 '17 at 20:04
  • In an earlier comment, which I deleted, I suggested that the Markov-Kakutani fixed point theorem could be applicable. I looked up one proof of that theorem, and it appears that the method suggested by Vogel plays an essential role; see John Conway, A Course in Functional Analysis, 2nd ed., section 5.10. – fred goodman Nov 06 '17 at 00:40
  • @Spaced: don't inf too early! Use the method to get $q(Ax - x) \le (1/n) p(A^n x - x) \le (1/n)[ p(A^nx) + p(-x)] \le (1/n) [p(x) + p(-x)]$ for all $n$. Hence $q(Ax - x) \le 0$, which is all that you need. – fred goodman Nov 06 '17 at 01:04
  • @fredgoodman Thanks for the reference and the advice. Thanks Vogel for showing me this approach. – Spaced Nov 06 '17 at 15:07