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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

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 $L^1(0,1)$.

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.

We use independent symmetric Rademacher random variables ${\epsilon }_j\in \{-1,1\}$, the intended meaning of the Bernoulli variables in this context. Without this convention the assertion need not hold: for one $0/1$ Bernoulli variable of success probability $t<1/2$ and one nonzero vector, the proposed inequality would be $\sqrt{t}\le \sqrt{2}t$, which is false.

Let $\Omega =\{-1,1{\}}^n$ with the uniform probability measure, put $S(\epsilon )={\sum }_j{\epsilon }_j{x}_j$, and set $F(\epsilon )=\Vert S(\epsilon )\Vert$. We prove the sharp estimate rather than assuming the Kahane-Khintchine inequality. If ${\epsilon }^{(j)}$ denotes $\epsilon$ with its $j$th sign reversed, defineThe Walsh characters $w_A(\epsilon )={\prod }_{j\in A}{\epsilon }_j$ form an orthonormal basis of scalar functions on this finite cube, and $\mathcal{L}{w}_A=-|A|w_A$. Since $F(-\epsilon )=F(\epsilon )$, its coefficients for odd $|A|$ vanish. Every nonconstant remaining character has $|A|\ge 2$. Expanding in this orthonormal basis gives the even-function spectral gap on a hypercubeAt each $\epsilon$, the Hahn-Banach theorem supplies a real norming linear functional $\varphi$ with $\Vert \varphi \Vert \le 1$ and $\varphi (S(\epsilon ))=F(\epsilon )$. For a complex Banach space use its underlying real space. Since $\Vert z\Vert \ge \varphi (z)$,Adding gives $\mathcal{L}F\ge -F$. Multiplying by the nonnegative $F$ yields $F(-\mathcal{L}F)\le F^2$. ThereforeThe constant is sharp: take $n=2$ and $x_1=x_2\ne 0$ in the scalar field. The random norm is $0$ or $2\Vert x_1\Vert$, each with probability one half.

A Banach space has cotype 2 if a constant $C$, independent of the length of the family, satisfiesFor $f_j\in L^1(0,1)$, the triangle inequality for the Euclidean norm givesFor real or complex scalars $a_j$, independence gives $\mathbb{E}|{\sum }_j{\epsilon }_j{a}_j{|}^2={\sum }_j|a_j{|}^2$. The proved sharp Rademacher second-moment inequality, applied pointwise to $(f_j(t))$, and Fubini theorem now yieldThus $L^1(0,1)$ has cotype 2, with $C\le \sqrt{2}$.

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 $Z\cong E\oplus W$, $Z\cong {\ell }^p$.

Use the finite-dimensional complex-module setting of the course. We prove the Weyl complete reducibility theorem by Haar averaging produces invariant complements, defining its constructions rather than invoking complete reducibility itself.

Choose a compact real form $\mathfrak{k}\subset \mathfrak{g}$: it is a real Lie subalgebra with $\mathfrak{g}=\mathfrak{k}\oplus i\mathfrak{k}$, and its simply connected Lie group $K$ is compact. Existence of this form and compact group is a standard fact for complex semisimple Lie algebras. Restrict the representation to $\mathfrak{k}$; integration of a Lie-algebra representation gives a group representation $R:K\to GL(V)$. Since $K$ is connected and generated by exponentials, the $\mathfrak{g}$-invariant subspace $W$ is also $K$-invariant.

Choose any positive-definite Hermitian inner product on $V$. With the Haar measure of $K$ normalized to mass one, defineThis is a Hermitian inner product, and it is positive-definite: for $v\ne 0$ every $R(k)v$ is nonzero, so the continuous positive integrand has positive integral. Right invariance of Haar measure gives $⟨R(a)v,R(a)w{⟩}_K=⟨v,w{⟩}_K$ for $a\in K$.

Let $W^{\prime }=W^{\perp }$ for this averaged inner product. Finite-dimensional inner-product linear algebra gives $V=W\oplus W^{\prime }$. If $v\in W^{\prime }$ and $w\in W$, thenbecause $R(a^{-1})w\in W$. Thus $W^{\prime }$ is $K$-invariant. Differentiating makes it $\mathfrak{k}$-invariant, and since it is a complex vector subspace, $\mathfrak{g}=\mathfrak{k}+i\mathfrak{k}$ also preserves it. ThereforeThe zero or whole-space submodule cases are included. This proves the required complement theorem without circularly assuming it.

The finite-dimensional hypothesis is essential to the statement. The nonsplit zero-weight Verma extension for sl2 is a counterexample for unrestricted modules: $M(0)$ has basis $v_n$ with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ has trivial one-dimensional quotient. A complement would be trivial and contained in the zero-weight line $\mathbb{C}{v}_0$, yet $f{v}_0=v_1\ne 0$. Hence no such complement exists in that infinite-dimensional case. The paper's finite-dimensional representation setting supplies the necessary hypothesis.