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Here “subspace” is understood to mean a closed closed linear subspace. Closedness matters: the finitely supported sequences $c_{00}$ are an infinite-dimensional algebraic subspace of ${\ell }^p$, but contain no infinite-dimensional complete subspace. Indeed, any such subspace would be the countable union of its intersections with the finite-dimensional coordinate spaces, contrary to the Baire category theorem. We prove the assertion with the intended closedness convention, for $1\le p<\infty$.

Write $E={\ell }^p$, the l-p sequence space. Choose positive numbers ${\delta }_n$ with ${\sum }_{n}2{\delta }_n<1$. Starting with $m_0=0$, choose $x_n\in X$ with $\Vert x_n{\Vert }_p=1$ and its first $m_{n-1}$ coordinates zero. This is possible because the kernel of finitely many coordinate linear functionals on an infinite-dimensional vector space is still infinite-dimensional. Choose $m_n>m_{n-1}$ so that the part of $x_n$ after coordinate $m_n$ has norm below ${\delta }_n$. Let $v_n$ be its restriction to $I_n=(m_{n-1},m_n]$, and put $u_n=v_n/\Vert v_n{\Vert }_p$. The triangle inequality givesThe vectors $u_n$ form a normalized block basic sequence. Their supports are disjoint, soConsequently $U=\overline{\mathrm{span}}\{u_n\}$ is isometric to the l-p sequence space.

For each block choose a norm-one linear functional ${\varphi }_n$ supported on $I_n$ with ${\varphi }_n(u_n)=1$. The Hahn-Banach theorem supplies such a functional, including when $p=1$. DefineThe series converges in the l-p sequence space, sinceAlso $Q{u}_n=u_n$, so $Q$ is a bounded projection onto $U$. Now define the bounded linear operatorThis series converges in operator norm and $\Vert A-I\Vert \le {\sum }_{n}2{\delta }_n<1$. The Neumann series therefore makes $A$ invertible. Disjoint support gives $A{u}_n=x_n$. Thus $Y=AU=\overline{\mathrm{span}}\{x_n\}$ lies in the closed subspace $X$, is isomorphic to the l-p sequence space, and is the range of the bounded projection $AQ{A}^{-1}$ on the whole space $E$. This proves the required copy of ${\ell }^p$ is complemented in the ambient ${\ell }^p$, not just in $X$.

Now let $Z$ be an infinite-dimensional complemented subspace of $E$. The construction supplies $Y\subset Z$ with $Y\cong E$ and a bounded projection $R:E\to Y$. Its restriction to $Z$ gives the topological direct sumIf $P:E\to Z$ is a bounded projection, then $(I-R)P$ is a bounded projection onto $W$: its range lies in $W$, and it fixes every element of $W$. Hence $W$ is a complemented subspace of $E$, so $E\cong W\oplus V$ for some closed $V$.

We give the absorption argument in the Pełczyński decomposition method explicitly. Reindexing coordinates by a bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ yields $E\cong ({⨁}_{n\ge 1}E{)}_p$. Applying the fixed Banach space isomorphism $E\cong W\oplus V$ in every coordinate givesThe extra $W$ is absorbed by shifting the countably many $W$ summands. The maps are bounded with bounded inverses because the same two-space decomposition is used uniformly in each summand; we may equip each finite direct sum with its equivalent $p$-sum norm. Since $Z\cong E\oplus W$, $Z\cong {\ell }^p$.

The real Grothendieck inequality asserts that a universal constant $K_G$ satisfiesfor every finite real scalar matrix and vectors of norm at most one in a real Hilbert space. For complex scalars the corresponding scalar supremum is over phases of modulus at most one; a universal complex constant also exists. We prove both versions, without needing the optimal constant.

First, for unit vectors $u,v$ and a standard Gaussian vector $G$ on their finite-dimensional span,To see this, rotational symmetry makes the angle of the Gaussian vector uniform in the plane spanned by $u,v$. The two signs differ with probability $\arccos ⟨u,v⟩/\pi$. Subtracting twice this probability from one gives the identity; collinear cases follow directly. Zero Gaussian coordinates have probability zero.

Put $c=\log (1+\sqrt{2})$, so $\sinh c=1$ and $0<c<\pi /2$. In the Hilbert direct sum of odd tensor powers, formed with the Hilbert tensor product inner product, defineThe Hilbert tensor product satisfies $⟨u^{\otimes m},v^{\otimes m}⟩=⟨u,v{⟩}^m$. For unit $u,v$, both transformed vectors have squared norm ${\sum }_{k\ge 0}{c}^{2k+1}/(2k+1)!=\sinh c=1$, and the power series for sine givesFor any finite collection of these transformed vectors, use a standard Gaussian vector on their finite-dimensional span. The sign identity and $|c⟨u,v⟩|<\pi /2$ giveConsequently, writing $M$ for the scalar sign supremum,Vectors of norm below one are padded to unit vectors in additional orthogonal directions, using separate directions for the two families so their cross inner products remain unchanged. This proves the real Grothendieck inequality with $K_R=\pi /(2\log (1+\sqrt{2}))$.

For the complex Grothendieck inequality, rotate the total scalar sum to have nonnegative real part. Its real part is the sum of the two real bilinear expressions involving $\mathrm{Re}{a}_{ij}\mathrm{Re}⟨u_i,v_j⟩$ and $-\mathrm{Im}{a}_{ij}\mathrm{Im}⟨u_i,v_j⟩$. The real inner products are realized in the underlying real Hilbert space, rotating one family by $i$ for the imaginary part. Each real coefficient matrix has real sign supremum at most the phase supremum of the original complex matrix. Applying the real inequality twice gives the valid constant $K_C=2K_R$. In what follows let $K_G$ denote the appropriate real or complex constant.

Now let $T:L^1(0,1)\to H$ be a bounded linear operator into a Hilbert space, and write $M=\Vert T\Vert$. If $M=0$ there is nothing to prove. Approximate a finite family of functions by simple functions on a common disjoint measurable partition. For such a family, write $f_j={\sum }_k{a}_{jk}{1}_{A_k}$ and $m_k=|A_k|>0$. The vectorshave norm at most $M$. Choose unit vectors $u_j$ with $⟨T{f}_j,u_j⟩=\Vert T{f}_j\Vert$; zero images contribute zero and can be assigned any unit vector. With inner products linear in their first argument, the Grothendieck inequality givesOver the real field, phases here are signs. For any Banach space $E$, operator norm duality givesIndeed, interchange the two suprema and choose each phase to align the scalar values of the functional. Thus the required summing inequality holds for simple functions. Approximation in the Lp space $L^1$ passes it to arbitrary finite families, since both sides vary continuously under finite sums of $L^1$ errors. We obtainFinally let $R$ be the closed span of the Rademacher functions in $L^1(0,1)$. Independence, the scalar second moment, and Q2 give, for each finite scalar family,Completion therefore defines a Banach space isomorphism $S:{\ell }^2\to R$ with $\Vert S\Vert \le 1$ and $\Vert S^{-1}\Vert \le \sqrt{2}$. If a bounded projection $P:L^1\to R$ existed, $S^{-1}P:L^1\to {\ell }^2$ would be an absolutely summing operator by the result just proved. The ideal property ${\pi }_1(BTA)\le \Vert B\Vert {\pi }_1(T)\Vert A\Vert$, obtained directly from the defining weak 1-norm, would make $I_{{\ell }^2}=(S^{-1}P)S$ absolutely summing. But its first $n$ coordinate vectors satisfyA single finite summing constant cannot dominate $\sqrt{n}$ for all $n$. Hence the closed Rademacher span is not complemented in $L^1(0,1)$.

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 a Banach space $E$ satisfying $E\cong ({⨁}_{n\ge 1}E{)}_p$, a complemented subspace $W$ is absorbed by $E$: write $E\cong W\oplus V$, expand the countable sum, and absorb one additional copy of $W$ into the countably many $W$ summands. Hence $E\oplus W\cong E$. If $Z$ is complemented in $E$ and contains a complemented copy of $E$, write $Z\cong E\oplus W$ with $W$ complemented in $E$; the preceding argument gives $Z\cong E$. The decompositions are Banach space isomorphisms, not merely algebraic decompositions.