On the proof of an invariant Hahn-Banach theorem

Question

In Rudin's book Functional Analysis there is something I do not understand in the proof of the following theorem (theorem 5.24 in this book):

Suppose $Y$ is a subspace of a normed linear space $X$, $f \in Y^*$, $\Gamma \subset \mathcal{B}(X,X)$, and that (a) $T(Y) \subset Y$ and $ST=TS$ for all $S,T \in \Gamma$, (b) $f \circ T = f$, for every $T \in \Gamma$. Then there exists $F \in X^*$ such that $F=f$ on $Y$, $||F||=||f||$, and $F \circ T = F$ for every $T \in \Gamma$.

In the proof one assumes that $||f||=1$, without loss of generality. Then the proof defines: $K= \lbrace \Lambda \in X^* : || \Lambda|| \leq 1\, , \, \Lambda = f\ \mathrm{on}\ Y \rbrace$

I understand that $K$ is a nonempty, convex, and weak*-compact subset of $X^*$.

The proof then defines a family of maps $\mathcal{F} = \lbrace \psi_T : T \in \Gamma \rbrace$ from $K$ to $X^*$, given by $ \psi_T(\Lambda)= \Lambda \circ T$.

I understand that $\mathcal{F}$ is a commuting family, and that the maps $\psi_T$ are affine, weak*-continuous, and that $\psi_T=f$ on $Y$.

What I do not understand, is that the proof claims that the maps $\psi_T$ are maps from $K$ into $K$. For this to be true, $||\Lambda \circ T||$ must be less or equal to 1 for all $T \in \Gamma$ and all $\Lambda \in K$. I do not understand this.

The claim seems wrong. Probably you can find a counterexample in $\mathbb R^2$. $X=\mathbb R^2, Y=\mathbb R\oplus\{0\}$ and $\Gamma=id \oplus \mathbb R$ may be such counterexample. — anything
– anything, Commented 11 hours ago
Thanks! Yes it seems, with your counterexample, that without any additional restrictions on $\Gamma$ in the statement of the theorem, the claim must be wrong. The proof of the theorem aims at using the Markov-Kakutani fixed-point theorem, but in order to use that theorem $\mathcal{F}$ should be a commuting family of continuous affine maps taking $K$ to itself. — Lars Nils
– Lars Nils, Commented 10 hours ago

Jochen Wengenroth · Accepted Answer · 2025-12-19 12:40:22Z

It seems that this a very rare case of a forgotten hypothesis in Rudin's books, you should indeed add that all $\Lambda\in\Gamma$ are contractions (which is the case in the application 5.25 of the theorem).

Here is a simple counterexample to the theorem as stated (it comes from a discussion with Leonhard Frerick):

Take $X=\mathbb R^2$ with the euclidean norm, $Y=\mathbb R\times\{0\}=\{se_1:s\in\mathbb R\}$ (with the first unit vector $e_1$) and $Z=\{tz:t\in\mathbb R\}$ a line through the origin which is not perpendicular to $Y$ so that the linear projection $\pi:X\to Y$, $se_1+tz\mapsto se_1$ has norm strictly bigger than $1$. Set $f(se_1)=s$, $T(se_1+tz)=se_1+2tz$, and $\Gamma=\{T\}$.

Then $f\circ T=f$ on $Y$ and, for every $T$-invariant extension $F\in X^*$, we have $F(z)=F(2z)=2F(z)$ so that $F(z)=0$. Choosing $s,t\in\mathbb R$ such that $\|se_1+tz\|>\|\pi(se_1+tz)\|=|s|$ we get $$|F(se_1+tz)|=|F(s)|=|s|<\|se_1+tz\|$$ which yields $\|F\|>1=\|f\|$.

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