# Approximating the norm of a finite dimensional representation on a Banach space by irreducible representations

Let $G$ be a compact group, let $X$ be a Banach space and let $\pi$ be a linear and isometric representation of $G$ on $X$ that is continuous with respect to the strong operator norm. For $v \in X$, we say that $v$ has a finite type if $span (\pi (G).v)$ is of finite dimension. It was shown by Shiga (in "Representations of a compact group on a Banach space") that vectors of finite type are dense in $X$.

My question is - when is the norm of subspaces of finite dimension in $X$ uniformly equivalent to some $l_q$-norm of their irreducible decomposition. Before making this precise, let me give two motivating examples:

Example 1 - $X=H$ a Hilbert space. In this case by Peter-Weyl theorem, for any finite dimensional subspace $V$ of finite type, we can write $V = V_1 \oplus ... \oplus V_n$, where $(V_1,\pi),...,(V_n, \pi)$ and the sum $V_1 \oplus ... \oplus V_n$ is taken with the $l_2$-norm. In other words, for every $v \in V$ there are $v_1 \in V_1,...,v_n \in V_n$ such that $$\Vert v \Vert^2 = \Vert v_1 \Vert^2+...+\Vert v_n \Vert^2.$$ Example 2 - $X$ is isomorphic to a Hilbert space, i.e., there is an invertible linear operator $T: X \rightarrow H$, such that $T,T^{-1}$ are bounded. In this case, for any subspace $V \subseteq X$ of finite dimension there are $(V_1,\pi) ,...,(V_n,\pi)$ irreducible representations such that for all $v \in V$, there are $v_1 \in V_1,...,v_n \in V_n$ such that $$\dfrac{1}{\Vert T \Vert \Vert T^{-1} \Vert} (\Vert v_1 \Vert^2+...+\Vert v_n \Vert^2) \leq \Vert v \Vert^2 \leq \Vert T \Vert \Vert T^{-1} \Vert (\Vert v_1 \Vert^2+...+\Vert v_n \Vert^2).$$ Let me sketch the proof in this case: define $\pi' = T \circ \pi \circ T^{-1}$ a representation of $G$ on $H$. We can make $\pi'$ into a unitary representation by changing the inner product of $H$ in the usual way, i.e., if $\langle . , . \rangle$ is the original inner product of $H$, we can define $\langle . , . \rangle_0$ as $$\langle u , v \rangle_0 = \int_{G} \langle \pi' (g). u , \pi ' (g). v \rangle d \mu (g) ,$$ where $\mu$ is the normalized Haar measure on $G$. Now let $V \subseteq X$ be a finite dimensional subspaces of $X$, then $TV$ is a finite dimensional subspace of $H$ and therefore there is a decomposition to irreducible representations $TV = V_1' \oplus ... \oplus V_n'$. We'll take $V_i = T^{-1} V_i'$ and this will do the trick. Indeed, let $v \in V$, then there are $v_1 \in V_1,...,v_n \in V_n$, with $$\Vert Tv \Vert^2_0 = \Vert T v_1 \Vert^2_0 + ... + \Vert T v_n \Vert^2_0$$ Notice that $$\Vert Tv \Vert^2_0 = \int_G \Vert \pi' (g). Tv \Vert^2 d\mu (g) = \int_G \Vert T \pi (g). v \Vert^2 d\mu (g),$$ and $$\int_G \Vert T \pi (g). v \Vert^2 d\mu (g) \leq \Vert T \Vert^2 \int_G \Vert \pi (g). v \Vert^2 d\mu (g) = \Vert T \Vert^2 \int_G \Vert v \Vert^2 d\mu (g) =\Vert T \Vert^2 \Vert v \Vert^2,$$ $$\int_G \Vert T \pi (g). v \Vert^2 d\mu (g) \geq \frac{1}{\Vert T^{-1} \Vert^2} \int_G \Vert \pi (g). v \Vert^2 d\mu (g) = \frac{1}{\Vert T^{-1} \Vert^2} \Vert v \Vert^2.$$ Therefore $$\frac{1}{\Vert T^{-1} \Vert^2} \Vert v \Vert^2 \leq \Vert Tv \Vert^2_0 \leq \Vert T \Vert^2 \Vert v \Vert^2.$$ The same inequalities hold for every $v_i$ and therefore we get the needed inequality.

After all this, here is my question: is there a condition on $X$, that is weaker then being isomorphic to a Hilbert space, that will insure that there is a constant $K>0$ and a set $Y$ of finite type vectors in $X$, such that $Y$ is dense in $X$ and for every $v \in Y$, $span (\pi (G).v)$ decomposes into a sum of irreducible representations $(V_1,\pi),..., (V_n,\pi)$ such that there are $v_1 \in V_1,..., v_n \in V_n$, such that for some $1 \leq q \leq \infty$ and some $K_1 >0, K_2 >0$ that may depend on $v$, we have that $K_1 K_2 \leq K$ and $$\dfrac{1}{K_1} (\Vert v_1 \Vert^q+...+\Vert v_n \Vert^q)^{\frac{1}{q}} \leq \Vert v \Vert \leq K_2 (\Vert v_1 \Vert^q+...+\Vert v_n \Vert^q)^{\frac{1}{q}}.$$ (in the case where $q = \infty$, this reads $$\dfrac{1}{K_1} \max (\Vert v_1 \Vert,...,\Vert v_n \Vert) \leq \Vert v \Vert \leq K_2 \max (\Vert v_1 \Vert,...,\Vert v_n \Vert).$$)

Notice that I want the constant $K$ to be universal for all $v \in Y$, but $q, K_1, K_2$ may be different for different $v$'s.

Edit: In response to Yemon Choi's remark - here are the cases that interest me:

1. Best case scenario (seems unlikely) - this holds for any $X$ of type >1.

2. More reasonable - this holds for $X$ given conditions on type and cotype.

3. Even more reasonable - one can say something for all $l^p$ spaces or maybe for some with suitable values of $p$ (I am more interested in the case where $p>2$).

Edit: The original question referred only to the $l_2$ norm, but I realized I don't need that restriction and can work with any $l_q$-norm.

• Given how strange Banach spaces can be, could you give us some idea of what kinds of $X$ you are intending to consider? For instance, are you hoping that this will work for $\ell_p$ for some $1<p<2$? – Yemon Choi Jun 17 '15 at 19:07
• Thanks for the comment - I edited the question in order to be more specific. – Izhar Oppenheim Jun 17 '15 at 19:17
• A silly answer: I guess a generic Banach space has no isometry except the multiplication by scalars of modulus one. For these spaces what you want obviously holds... – Mikael de la Salle Jun 17 '15 at 23:08
• More serioulsy, a space of the form $L^2(\{0,1\}^{\mathbf N};X)$ satisfies what you want if and only if $X$ is isomorphic to a Hilbert space (because for the abelian group $\{0,1\}^N$ this implies that X has same type and cotype $q$, so that $q=2$ and $X$ is a Hilbert space). This is an evidence that nothing interesting can be said outside of spaces isomorphic to Hilbert spaces. – Mikael de la Salle Jun 17 '15 at 23:17