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Two probability measures supported on one compact interval are equal if all their moments are equal. Indeed, equality on monomials gives equality on polynomials, the Weierstrass approximation theorem extends it to every continuous function on the interval, and compactly supported continuous functions determine a finite Borel measure.

Given a cost $c$ and probability measures $\mu ,\nu$, the Monge problem minimizes $\int c(x,T(x))d\mu (x)$ over measurable transport maps with pushforward measure $T_{\#}\mu =\nu$. The feasible set may be empty because a map cannot split an atom of a measure.

A non-atomic measure contains no atom of a measure. For a finite Borel measure on a Polish space, this is equivalent to giving every singleton mass zero. For a probability measure on $\mathbb{R}$, it is also equivalent to continuity of its cumulative distribution function.

On binary words of length $n$, normalized Hamming distance is the fraction of differing coordinates. Unlike unscaled Hamming distance, it makes the Hamming cubes a Lévy family under uniform probability measures.

Optimal transport minimizes the cost of moving one probability measure to another. The Monge optimal transport problem uses a transport map; the Kantorovich optimal transport problem allows a transport plan that can split mass.

The side of the inscribed equilateral triangle has length $\sqrt{3}r$. If $d$ is the distance of a chord midpoint from the center, a right triangle gives chord length $2\sqrt{r^2-d^2}$. Thus a chord is sufficiently long exactly when $d<r/2$.

(a) Uniform midpoint area puts the midpoint in the radius-$r/2$ disk with probability(b) Fix the first endpoint by rotational symmetry. Independence and uniformity make the angle $\theta$ of the second endpoint uniform on $[0,2\pi )$. Its chord length is $2r\sin (\theta /2)$, so the required event is $2\pi /3<\theta <4\pi /3$. Hence(c) If the midpoint distance itself has uniform distribution on $[0,r]$, the event $d<r/2$ has probabilityThese are three different probability measures on chords. Their different answers are Bertrand's paradox: specifying what is uniform is indispensable. Uniform area, uniform endpoints and uniform radial distance are not interchangeable sampling rules. Equality at the length threshold has probability zero in all three models.

Work in discounted units with a riskless asset of value one, trivial initial information and finitely many risky assets. Let $Y=S_1-S_0\in L^2(P;{\mathbb{R}}^d)$ be their discounted gains. A terminal attainable claim has the form $v+{\theta }^{T}Y$, whose initial cost is $v$. Their spanis a finite-dimensional closed subspace of a Hilbert space. For $H\in L^2(P)$, one-period least-squares hedging is its orthogonal projection onto $\mathcal{H}$.

Let $m=\mathbb{E}Y$, $\Sigma =\mathrm{Cov}(Y)$ and $b=\mathrm{Cov}(Y,H)$. Remove redundant risky directions so that $\Sigma$ is invertible. Under absence of arbitrage, a gain direction with zero variance is a constant and must be zero, so this reduction loses no genuine trading opportunity. For freely chosen capital and holdings, minimizingfirst gives $v=\mathbb{E}H-{\theta }^{T}m$. The remaining centered error isCompleting the square yieldsThe residual $L=H-v^{\ast }-({\theta }^{\ast }{)}^{T}Y$ satisfies $\mathbb{E}L=0$ and $\mathbb{E}[LY]=0$, the least-squares normal equations. Exact claim replication is possible precisely when this residual is zero. If capital is fixed at $v$, its normal equation instead gives the fixed-capital quadratic hedge in a one-period marketThus fixed-capital and freely optimized hedging are distinct problems.

A dominated martingale measure is a probability measure $Q\ll P$ with integrable gains and ${\mathbb{E}}_{Q}Y=0$. Its density $Z$ obeys $Z\ge 0$, $\mathbb{E}Z=1$, $\mathbb{E}[ZY]=0$. An equivalent martingale measure additionally has $Z>0$ almost surely. Every such measure prices an attainable discounted claim $v+{\theta }^{T}Y$ at $v$, but different measures may give different expectations to an unattainable claim.

The minimal martingale measure in a one-period market is defined by preserving the mean-zero martingale directions orthogonal to the gain innovation $Y-m$. Among square-integrable densities it isIndeed $\mathbb{E}{Z}_{\ast }=1$, $\mathbb{E}[Z_{\ast }Y]=m-\Sigma {\Sigma }^{-1}m=0$, and for $\mathbb{E}L=0$, $\mathbb{E}[L(Y-m)]=0$ we have $\mathbb{E}[Z_{\ast }L]=0$. Conversely preservation of all these orthogonal directions puts $Z$ in $\mathrm{span}\{1,Y-m\}$; normalization and zero gain expectations then determine $Z_{\ast }$. Any other square-integrable signed pricing density differs from $Z_{\ast }$ by a vector in ${\mathcal{H}}^{\perp }$, so $Z_{\ast }$ also has the smallest $L^2$ norm. Its evaluation of a claim isThis is the capital of the optimal quadratic hedge, not automatically an arbitrage-free price. The formula may define only a signed martingale measure: $Z_{\ast }$ is a genuine dominated measure only if it is nonnegative, and equivalent only if strictly positive. For example, gains $(-1,1,2)$ with probabilities $(1/10,4/5,1/10)$ have $m=9/10$, $\Sigma =49/100$ and $Z_{\ast }(2)=-50/49$. Yet probabilities $(11/20,7/20,1/10)$ are all positive and have zero gain mean. Thus absence of arbitrage does not force a positive minimal density.

For the market completeness assertion, assume absence of arbitrage and hence existence of an equivalent martingale measure $Q_0$. In finite states this is the usual fundamental theorem of asset pricing; the one-period version also holds with finitely many assets on a general probability space. Market completeness means that every bounded claim is attainable, equivalently $\mathcal{H}=L^2(P)$ in this square-integrable setting. We now prove completeness and uniqueness of dominated martingale measures.

If the model is complete, replicate every indicator $1_A$. Two dominated martingale measures both evaluate it at its unique claim replication cost, and therefore assign every event $A$ the same probability. They are identical.

Conversely let $k=\dim \mathcal{H}$ and suppose the model is incomplete. There exist $k+1$ disjoint events $A_j$ of positive probability. Otherwise the probability space would have at most $k$ atoms: any non-atomic positive event could be split to increase the number. Its whole payoff space would then have dimension at most $k$, forcing equality with $\mathcal{H}$ and market completeness. Choose a basis $h_1,\dots ,h_k$ of $\mathcal{H}$. Its members are integrable under $Q_0$, because they are combinations of the constant and the gains. The $k+1$ vectorsare linearly dependent. Hence a nonzero bounded $g={\sum }_j{a}_j{1}_{A_j}$ satisfies ${\mathbb{E}}_{Q_0}[gh]=0$ for every $h\in \mathcal{H}$, in particular $h=1,Y_i$. For $0<\epsilon <1/\Vert g{\Vert }_{\infty }$, defineThese strictly positive densities integrate to one, preserve every gain expectation, and give distinct equivalent martingale measures. They are also dominated by $P$. Thus uniqueness is impossible in an incomplete market, proving the equivalence under the stated no-arbitrage hypothesis.

That hypothesis is essential. On three positive-probability states, the gain $(0,1,2)$ has the unique dominated martingale measure, concentrated on the first state, but its attainable span has dimension only two. The market is incomplete and admits arbitrage. This explains why uniqueness among merely dominated probabilities must not be used without an equivalent pricing measure or a no-arbitrage assumption.

On a filtered probability space carrying a standard Brownian motion $W$ with independent increments relative to the filtration, a strictly positive security price follows geometric Brownian motion with constant drift $\mu$ and volatility $\sigma$ if it solves the stochastic differential equationHere $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and has the normal distribution $N(0,t-s)$ for $s<t$. The solution isIndeed applying the Itô formula to this exponential gives the specified drift and volatility. Consequently the price is positive and continuous, andThus disjoint log-return increments are independent and stationary. The parameter $\mu$ is the instantaneous expected relative return under the physical probability measure, not the mean log-return. Under the pricing risk-neutral measure it becomes the risk-free rate for a non-dividend-paying stock.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

Take bank-account values $B_k=R^k$ with $R>0$, and let the stock multiply independently by $u$ or $d$ at each step, where $0<d<R<u$. Under the physical probability measure $P$, the up probability is $p\in (0,1)$; the filtration records the successive moves. These assumptions specify the standard binomial market. If $R$ lies outside the strict interval $(d,u)$, a long or short stock position financed through the bank account gives an arbitrage.

The risk-neutral probability in a binomial market is fixed by $qu+(1-q)d=R$, givingFor successor claim values $V_u,V_d$ at a node with current stock price $S$, solve the two replication equations. The current stock holding and current cash value areThe cost of this replicating portfolio in a binomial market isApply this backward option pricing recursion from any terminal contingent claim $H$, including claims depending on the whole sequence of moves. It constructs a self-financing portfolio and proves market completeness by induction on the remaining number of periods. Taking successive conditional expectations under the measure $Q$ with independent up probability $q$ gives the unique risk-neutral pricing valueAny other price would give an arbitrage by buying the cheaper of the claim and its replica and selling the dearer.

If $K$ is the number of up moves, a particular path has probabilities $p^K(1-p{)}^{n-K}$ under $P$ and $q^K(1-q{)}^{n-K}$ under $Q$. Therefore the binomial-market probability density, namely the Radon-Nikodym derivative, isThis last expression makes $Z$ a function of the terminal stock price. Summing over paths gives ${\mathbb{E}}_{P}Z=1$, and strict positivity proves equivalence of the two probability measures.

For unrestricted terminal portfolio wealth $H$, its initial replication cost is $R^{-n}{\mathbb{E}}_{Q}H$. Hence feasible wealth satisfies the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. For the stated exponential utility, $v^{\prime }(x)=e^{-ax}$ and $v^{\prime \prime }(x)=-a{e}^{-ax}<0$. The binomial exponential-utility terminal wealth iswhereIt satisfies the budget, and its marginal utility is $v^{\prime }(H^{\ast })=\lambda Z$ with $\lambda =\exp (-a{R}^n{w}_0-{\mathbb{E}}_Q\log Z)>0$. The concave supporting-tangent inequality givesTaking expectations under $P$ makes the last term zero by the budget. Strict concavity makes equality possible only when $H=H^{\ast }$ in every positive-probability state. The already constructed complete market supplies its replicating strategy, so this proves global optimality and uniqueness. The formula allows negative final wealth, as permitted by the unrestricted trading problem.

Write $T=t_0$, $B_t=e^{-\rho (T-t)}$, and let $V_t=f(S_t,t)$ with $f(x,t)=xg(x,t)+B_{t}h(x,t)$. The portfolio holds $g(S_t,t)$ shares of stock and $h(S_t,t)$ units of the zero-coupon bond. It is self-financing when all changes of holdings are funded within it, which is the gain identityUnder the physical probability measure, write $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, where $\sigma >0$ and $d{B}_t=\rho B_{t}dt$. The Itô formula givesMatching the Brownian motion coefficients forces $f_x=g$. Since $f_x=g+x{g}_x+B_t{h}_x$, this is preciselyDifferentiating $f_x=g$ gives $f_{xx}=g_x$. Since $f_t=x{g}_t+\rho B_{t}h+B_t{h}_t$, matching the drift coefficients now givesThese are the self-financing conditions for smooth stock and bond holdings. Conversely, substituting both identities into the Itô formula gives the gain identity, so they are sufficient as well. Smoothness and the positive lognormal distribution density of $S_t$ extend the identities along the price process to every $x>0$, with boundary times obtained by continuity.

The portfolio with constant stock value in bond units must have $xg(x,t)=\theta B_t$, so $g(x,t)=\theta B_t/x$. The first condition then gives $h_x=\theta /x$, henceThe second condition reduces to $c^{\prime }(t)=\theta ({\sigma }^2/2-\rho )$. Initial portfolio wealth requires $h(S_0,0)=w_0/B_0-\theta =w_0{e}^{\rho T}-\theta$. Therefore the holdings areThe corresponding portfolio wealth isUnder the risk-neutral measure, $\log (S_t/S_0)=(\rho -{\sigma }^2/2)t+\sigma W_t^Q$, so its value in zero-coupon bond units is $V_t/B_t=w_0{e}^{\rho T}+\theta \sigma W_t^Q$. This also verifies the self-financing gain equation after using the bond as numéraire.

A standard form of Varadhan's integral theorem is as follows. Let ${\mu }_L$ be probability measures on a Polish space $E$ satisfying a large deviation principle with speed $L\to \infty$ and good rate function $I$. If $F:E\to \mathbb{R}$ is continuous and bounded above, thenThe usual bounded-continuous version follows immediately. If $F$ is unbounded above, the same conclusion holds under the additional exponential tail conditionThis is the exponential tail extension of Varadhan's lemma. A convenient sufficient hypothesis isAn integrability hypothesis is essential for general unbounded functions; the ordinary large deviation principle alone does not control arbitrarily large exponential weights.

For example, let $X^L=L$ with probability $e^{-L^2}$ and $X^L=0$ otherwise. It satisfies the large deviation principle at speed $L$ with rate zero at zero and infinite elsewhere. But for $F(x)=x^2$, $\mathbb{E}{e}^{LF(X^L)}\ge e^{L^3-L^2}$, whose logarithmic rate diverges, while $\sup (F-I)=0$. This shows why the tail hypothesis cannot simply be omitted.

For the lower bound, fix $x$ with $I(x)<\infty$ and $\epsilon >0$. By continuity choose an open set $G$ containing $x$ on which $F(y)\ge F(x)-\epsilon$. ThenApply the open-set lower bound of the large deviation principle:Let $\epsilon \downarrow 0$ and take the supremum over $x$ to obtain the required lower bound. This argument also applies when $F$ is not bounded above.

For the upper bound, suppose $F\le M$ and set $S={\sup }_E(F-I)$, which is finite. Choose a large $K$ and partition $[-K,M]$ into finitely many closed intervals $[a_j,a_j+\epsilon ]$ of length at most $\epsilon$. The preimagesare closed sets. The overlapping endpoints do not matter for an upper estimate. Split off the lower tail and bound each remaining exponential weight:For a finite sum of nonnegative terms the upper logarithmic rate is at most the largest of their upper logarithmic rates. Applying the closed-set upper bound givesFor $x\in C_j$, $a_j+\epsilon \le F(x)+\epsilon$, hence every inner term is at most $S+\epsilon$. Empty preimages contribute no term. Let $K\to \infty$ and then $\epsilon \downarrow 0$ to get the upper bound $S$. Together with the lower bound this proves the bounded-above theorem.

For the unbounded extension, put $F_M=\min (F,M)$. This is continuous and bounded above, so the proved result applies, withThe full integral is at most the integral with $F_M$ plus the integral on $\{F>M\}$. The upper logarithmic rate is therefore at most the larger of $S_M$ and the tail rate. The tail hypothesis and then $M\to \infty$ give the upper bound $S$; the earlier local argument gives the lower bound. If $S=+\infty$, that lower argument already yields the asserted infinite limit.

Finally, the sufficient exponential-moment condition really implies the required tail estimate: on $F>M$,After integration the upper logarithmic rate is bounded by a finite constant minus $(\gamma -1)M$, which tends to $-\infty$. This completes the statement and proof of Varadhan's lemma, including the hypotheses needed for the unbounded version.

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 $C\le \sqrt{2}$.

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 $N$.