= Solution
For an <orthonormal basis> $(e_i)$, define the <Hilbert-Schmidt norm> by $\|T\|_2^2=\sum_i\|Te_i\|^2$. For a possibly uncountable <basis>, the nonnegative sum means the supremum over finite subsets. <Parseval's identity> gives, for any other <orthonormal basis> $(f_j)$,
$$
\sum_i\|Te_i\|^2=\sum_{i,j}|\langle Te_i,f_j\rangle|^2
=\sum_j\|T^*f_j\|^2.
$$
Changing the domain <basis> in the last expression proves <basis> independence. Polarization gives a basis-independent <inner product>
$$
\boxed{\langle T,S\rangle_2=\sum_i\langle Te_i,Se_i\rangle
=\operatorname{Tr}(S^*T),}
$$
with the same first-argument linear convention. The sums converge by the <Cauchy-Schwarz inequality>. The operator-norm bound $\|T\|\leq\|T\|_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit <vector>.
To prove completeness, let $(T_n)$ be Cauchy in this <norm>. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the <basis>,
$$
\sum_{i\in F}\|(T_n-T)e_i\|^2
=\lim_{m\to\infty}\sum_{i\in F}\|(T_n-T_m)e_i\|^2.
$$
For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is <Hilbert-Schmidt> and $\|T_n-T\|_2\to0$; it also shows $T$ is <Hilbert-Schmidt>. Thus \b[the <Hilbert-Schmidt operators> form a <Hilbert space>]. <Finite-rank operators> are dense there: keeping finitely many <basis> columns makes the sum of omitted squared column <norms> tend to zero. The operator-norm bound then shows that every <Hilbert-Schmidt operator> is compact.
Now let $U$ be a strongly continuous <unitary representation> of the compact metric <group> $G$, as is standard for a representation of a topological <group>. Fix $\xi\ne0$ and average the positive <rank-one operator> $R_{\xi,\xi}$:
$$
K=\int_G U_gR_{\xi,\xi}U_g^*\,dg.
$$
The integrand is continuous in the <Hilbert-Schmidt norm>, so this <Bochner integral> exists in the complete space just proved. Normalized <Haar measure> and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,
$$
\langle K\xi,\xi\rangle=\int_G|\langle U_g\xi,\xi\rangle|^2\,dg>0:
$$
the integrand is positive near the identity and every nonempty open set has positive <Haar measure>. Thus $K\ne0$. A nonzero <eigenspace> is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero <invariant subspace> of smallest dimension; it is irreducible. Its <orthogonal complement> is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.
Take a maximal orthogonal family of such finite-dimensional irreducible <invariant subspaces>. If their closed span had a nonzero <orthogonal complement>, the same averaging argument on that complement would produce another member, contradicting maximality. Therefore
$$
\boxed{H=\bigoplus_\alpha H_\alpha,
\quad\dim H_\alpha<\infty,\quad U|_{H_\alpha}\text{ irreducible}.}
$$
This <compact averaging of a rank-one operator> argument does not require $H$ itself to be separable. Strong <continuity> is the topological representation hypothesis used in the averaging step.
Back to article page
