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An absolutely summing operator is an absolutely p-summing operator with $p=1$. The identity on the infinite-dimensional l2 sequence space is not absolutely summing: the first $n$ coordinate vectors have sum of norms $n$, while their weak 1-norm is $\sqrt{n}$.

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.

For Banach spaces $E,F$ and $1\le p<\infty$, define the weak $p$-norm of a finite family byA bounded linear operator $T:E\to F$ is an absolutely p-summing operator if $({\sum }_i\Vert T{x}_i{\Vert }^p{)}^{1/p}\le C{w}_p(x_1,\dots ,x_m)$ for every finite family. The least constant is ${\pi }_p(T)$.

The Pietsch factorization theorem states that such an operator admits a probability measure $\mu$ on $K=B_{E^{\ast }}$ with its weak-star topology such thatEquivalently, let $i:E\to C(K)$ be the evaluation embedding $i(x)(\varphi )=\varphi (x)$ and $j_p:C(K)\to L^p(\mu )$ the natural map. There is a bounded linear operator $V:H\to F$, where $H=\overline{j_{p}i(E)}\subset L^p(\mu )$, such that $T=V{j}_{p}i$ and $\Vert V\Vert \le {\pi }_p(T)$. Conversely, any such factorization makes $T$ an absolutely p-summing operator, with ${\pi }_p(T)\le \Vert V\Vert$. The final operator is required on this closed subspace $H$; no extension to all of $L^p(\mu )$ is assumed.

Here is the existence proof. If $T=0$ the conclusion is immediate. Otherwise write $C={\pi }_p(T)>0$. The Banach-Alaoglu theorem makes $K$ compact, and $\varphi \mapsto |\varphi (x){|}^p$ is continuous for each $x\in E$. For finitely many vectors $x_1,\dots ,x_m$, putWe claim the compact convex hull of $v(K)$ meets $a+[0,\infty {)}^m$. If it did not, finite-dimensional Hahn-Banach separation theorem would give weights ${\lambda }_i\ge 0$ withThe weights are nonnegative because the second convex set is unbounded in every positive coordinate direction. This strict inequality contradicts the defining summing inequality applied to ${\lambda }_i^{1/p}{x}_i$. Thus a finite convex combination of the vectors $v(\varphi )$ lies coordinatewise above $a$. Equivalently, a finitely supported probability measure satisfies all these finitely many domination constraints.

By the Riesz-Markov-Kakutani representation theorem, probability measures on compact $K$ are the positive norm-one functionals on $C(K)$ taking the constant function to one. They form a compact set in the weak-star topology, by the Banach-Alaoglu theorem. Each domination constraint $\int |\varphi (x){|}^{p}d\mu \ge C^{-p}\Vert Tx{\Vert }^p$ is closed. The finite feasibility just proved gives the finite intersection property, hence one measure $\mu$ satisfies every constraint.

Define $V(j_{p}i(x))=Tx$. The domination inequality makes this well-defined and bounded by $C$, and completeness of $F$ extends it to $H$. Conversely, any factorization satisfiesThis proves both directions of the Pietsch factorization theorem, including the optimal factorization norm.

For the multiplication operator, let $f\in L^p(0,1)$ and take $g_1,\dots ,g_m\in C([0,1])$. Point evaluations are norm-one linear functionals on $C([0,1])$, soThus $M_f$ is an absolutely p-summing operator and ${\pi }_p(M_f)\le \Vert f{\Vert }_p$. Taking the single function $g=1$ gives the reverse inequality, soFor $f\ne 0$, the explicit domination measure on $[0,1]$ is $|f(t){|}^{p}dt/\Vert f{\Vert }_p^p$, pushed into $K$ by $t\mapsto {\delta }_t$. For $f=0$ everything is zero; no essential boundedness of $f$ is needed.

Let $C$ be a cotype 2 constant of $Y$. All bounded linear operators from ${\ell }_{\infty }^N$ have finite p-summing norms: writing $Tx={\sum }_{k=1}^N{x}_k{y}_k$ and using the triangle inequality in the finite-family ${\ell }^p$ norm gives ${\pi }_p(T)\le {\sum }_k\Vert y_k\Vert$. Thus applying the Pietsch factorization theorem below is legitimate.

First let $a_4={\pi }_4(T)$, and use the Pietsch factorization theorem to choose a probability measure $\mu$ on the dual unit ball $B_{{\ell }_1^N}$. For a finite family $x_j\in {\ell }_{\infty }^N$, the cotype 2 inequality and Pietsch factorization theorem domination giveThe Jensen inequality bounds the final bracket by $[\int \mathbb{E}|{\sum }_j{\epsilon }_j\varphi (x_j){|}^{4}d\mu {]}^{1/4}$. For real or complex scalars,For complex scalars, paired-index expansion gives $\mathbb{E}|{\sum }_j{\epsilon }_j{b}_j{|}^4=2({\sum }_j|b_j{|}^2{)}^2+|{\sum }_j{b}_j^2{|}^2-2{\sum }_j|b_j{|}^4$, which is at most the displayed bound by the triangle inequality. For real scalars this reduces to $3(\sum b_j^2{)}^2-2\sum b_j^4$. Since $\mu$ is a probability measure on the dual unit ball, we obtainTaking the least constant proves ${\pi }_2(T)\le K{\pi }_4(T)$ with $K=3^{1/4}C$, uniformly in $N$.

We next prove a summing-norm interpolation estimate using coordinate truncation. Put $a={\pi }_2(T)$ and $M=\Vert T\Vert$, assuming $T\ne 0$. The Pietsch factorization theorem for $p=2$ gives a measure $\mu$ on $B_{{\ell }_1^N}$. The Cauchy-Schwarz inequality givesSet ${\nu }_i=\int |{\varphi }_i|d\mu$. Their sum is at most one; adding the missing mass to one coordinate, if necessary, makes $({\nu }_i)$ a probability vector and preserves domination. HenceWrite $r=({\sum }_i{\nu }_i|x_i{|}^4{)}^{1/4}$. For $t>0$, clip each coordinate of $x$ at magnitude $t$, obtaining $z_i=x_i\min (1,t/|x_i|)$, with $z_i=0$ if $x_i=0$. Set $w=x-z$. Then $\Vert z{\Vert }_{\infty }\le t$ andThe triangle inequality and the two different bounds on $T$ now give $\Vert Tx\Vert \le Mt+a{r}^2/t$. For $r>0$, take $t=r\sqrt{a/M}$; if $r=0$, domination already gives $Tx=0$. ConsequentlySumming fourth powers over any family $(x_j)$ givessince each coordinate evaluation belongs to the dual unit ball. Therefore ${\pi }_4(T)\le 2\sqrt{M{\pi }_2(T)}$. Combine this with the first estimate:The zero operator also satisfies this inequality. Both constants depend only on $Y$, through its cotype 2 constant, and not on $N$.

Finally let $E\subset Y$ have dimension $N$, with its inherited norm. The permitted finite-dimensional identity result is ${\pi }_2(I_E)=\sqrt{N}$. The inclusion $J:E\to Y$ is an isometric embedding, so ${\pi }_2(J)={\pi }_2(I_E)$. For every Banach space isomorphism $S:{\ell }_{\infty }^N\to E$, apply the preceding estimate to $T=JS$ and the ideal property of absolutely p-summing operators:The Banach-Mazur distance is the infimum of this last product over $S$, and therefore

For an absolutely p-summing operator $T:E\to F$, there is a probability measure $\mu$ on the dual unit ball with its weak-star topology such that $\Vert Tx\Vert \le {\pi }_p(T)(\int |\varphi (x){|}^{p}d\mu (\varphi ){)}^{1/p}$. Thus $T$ factors through the closed span of the evaluation functions in $L^p(\mu )$, with the final operator having norm at most ${\pi }_p(T)$. Conversely this factorization implies the summing inequality by summing under the integral. Existence of $\mu$ follows from finite-dimensional Hahn-Banach separation theorem: failure for finitely many vectors would contradict the summing inequality for a nonnegative weighted family. Compactness of probability measures then combines the finite constraints. The final operator need only be defined on the indicated closed subspace, not on all of $L^p(\mu )$.