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For a bounded linear operator $T:E\to F$, write the weak square norm of a finite family asIt is a 2-summing operator if $({\sum }_j\Vert T{x}_j{\Vert }^2{)}^{1/2}\le C{w}_2(x_1,\dots ,x_m)$ for all finite families, and ${\pi }_2(T)$ is the least such $C$. This specializes the definition of an absolutely p-summing operator.

First let $E=H_1$, $F=H_2$ be Hilbert spaces. For the column operator $A:{\ell }_2^m\to H_1$, $A{e}_j=x_j$, the Riesz representation theorem gives $w_2(x_1,\dots ,x_m)=\Vert A^{\ast }\Vert =\Vert A\Vert$. If $T$ is a Hilbert-Schmidt operator, the Parseval identity applied to orthonormal bases givesThus ${\pi }_2(T)\le \Vert T{\Vert }_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the Bessel inequality, with equality by choosing one of its vectors. Hence ${\sum }_j\Vert T{e}_j{\Vert }^2\le {\pi }_2(T{)}^2$. Taking the supremum over finite subsets of an orthonormal basis proves the Hilbert-Schmidt operator condition and the reverse norm inequality. These basis sums work even for nonseparable Hilbert spaces, being understood as suprema of finite subsums. Therefore

For an $n$-dimensional normed vector space $E$, again associate a family $(x_j)$ with $A:{\ell }_2^m\to E$. By the Hahn-Banach theorem and scalar Euclidean duality,Let $Q$ be the orthogonal projection onto $(\ker A{)}^{\perp }$, whose dimension is at most $n$. Since $A=AQ$, the trace of $Q$ givesThis proves ${\pi }_2(I_E)\le \sqrt{n}$ for every norm, without comparing that norm to an arbitrary Euclidean norm.

To obtain equality and the projection bound, choose a maximum-volume centered ellipsoid inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the John ellipsoid; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in \partial B_E\cap \partial B_2$ and positive weights $c_r$ with the John contact decompositionFor completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\mathrm{tr}D>0$ and $u^{\ast }Du\le 0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\mathrm{tr}D+O(t^2)$ in the real case and $|{\det }_{\mathbb{C}}(I+tD){|}^2=1+2t\mathrm{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.

At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is ${\varphi }_r(x)=u_r^{\ast }x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|{\varphi }_r(x)|\le \Vert x{\Vert }_E$. Also $B_2\subseteq B_E$ gives $\Vert x{\Vert }_E\le |x{|}_2$. For any dual-unit functional $\varphi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}{u}_r$:The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence ${\pi }_2(I_E)\ge \sqrt{n}$. Together with the upper estimate, the norm isThis proves the 2-summing norm of a finite-dimensional identity formula.

Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each ${\varphi }_r$ to a dual-unit functional ${\tilde{\varphi }}_r$ on $X$ using the real or complex Hahn-Banach theorem. DefineThe contact identity gives $P{|}_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto {\sum }_r\sqrt{c_r}{a}_r{u}_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. ThereforeThus the linear projection satisfies $\boxed{\Vert P\Vert \le \sqrt{n}}$, the Kadec-Snobar projection bound. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.