Bourbaki

Exercise 27, § 5

Tag 0382, permanent. Cite it as [Bourbaki, Tag 0382].

Spectral Theories, IV, p. 364

Let M=X2Y2(X2+Y2−3)∈R[X,Y]M = X^2Y^2(X^2+ Y^2-3)\in \mathbf{R}[X,Y] ("Motzkin polynomial"). (a) One has M(x,y)⩾0M(x, y)\geqslant 0 for all xx and yy in R\mathbf{R}.

(b) There exists no finite family (pi)i∈I(p_i)_{i\in I} in R[X,Y]\mathbf{R}[X,Y] such that M=∑pi2M =\sum p^2_i. (If such a family existed, now prove that each pip_i must be of the form a+ba+bXY with a∈Ra\in \mathbf{R} and b∈R[X,Y]b\in \mathbf{R}[X,Y] of degree at most 1; then calculate the coefficient of (XY)2.^2.)

(c) Let τ:N2→N−{0}\tau :\mathbf{N}^2\rightarrow \mathbf{N}-\{0\} be the mapping defined by

τ(0,0)=1,τ(1,2)=2,τ(2,1)=3,τ(1,1)=4,τ(1,0)=5 \tau (0,0) = 1,\tau (1,2) = 2,\tau (2,1) = 3,\tau (1,1) = 4,\tau (1,0) = 5
τ(0,1)=6,τ(2,0)=7,τ(0,2)=8,τ(3,0)=9,τ(0,3)=10 \tau (0,1) = 6,\tau (2,0) = 7,\tau (0,2) = 8,\tau (3,0) = 9,\tau (0,3) = 10

and

τ(i,j)=j+1+(i+j)(i+j+1)/2\tau (i, j) =j+ 1 + (i+j)(i+j+ 1)/2 if i+j⩾4i+j\geqslant 4.

The mapping τ\tau is a bijection.

(d) Let (νn)n⩾1(\nu_n)_{n\geqslant 1} be the sequence defined by

νi=1\nu_i= 1 if i∈{1,2,3},ν4=4,νn=(ni\in \{1,2,3\},\nu_4= 4,\nu_n= (n!)(n+1)!^{(n+1)!} if n⩾5n\geqslant 5.

For nn and mm in N\mathbf{N}, put μn,m\mu_{n,m} = 0 if nn or mm is odd and μn,m=ντ(n/2,m/2)\mu_{n,m}=\nu_{\tau(n/2,m/2)} otherwise. The linear mapping α\alpha of C[X,Y]\mathbf{C}[X,Y] into C\mathbf{C} such that α(XnYm)=μn,m\alpha (X^nY^m) =\mu_{n,m} satisfies α(p2)⩾0\alpha (p^2)\geqslant 0 for every p∈R[X,Y]p\in \mathbf{R}[X,Y]. (It is enough to verify that, for every integer k∈Nk\in \mathbf{N}, the determinant of the matrix (am,m)1⩽n,m⩽k(a_{m,m})_{1\leqslant n,m\leqslant k} is positive, where an,ma_{n,m} is defined by aτ(i,j),τ(i′,j′)=α(Xi+i′Yj+j′)a_{\tau(i,j),\tau(i',j')}=\alpha (X^{i+i'}Y^{j+j'}) for all (i,j,i′,j′)(i, j, i', j'); then proceed by induction on k.k.)

(e) There exists no positive measure ν\nu on R2\mathbf{R}^2 such that (x,y)↦xnym(x, y)\mapsto x^ny^m is μ\mu-integrable for all (n,m)∈N2(n, m)\in \mathbf{N}^2, and such that

∫R2xnymdν(x,y)=μn,m \int_{\mathbf{R}^2}x^ny^md\nu (x, y) =\mu_{n,m}

for all (n,m)∈N2(n, m)\in \mathbf{N}^2. (f) There exist a Hilbert space E and a partial operator uu with dense domain on E satisfying the following conditions:

(i) One has dom(uu)⊂\subset dom(u∗u^*);

(ii) One has ∥u(x)∥=∥u∗(x)∥\|u(x)\|=\|u^*(x)\| for every x∈x\in dom(uu);

(iii) There exist no Hilbert space F containing E as a subspace, such that the canonical injection of E into F is isometric, and no normal partial operator vv on F such that u⊂vu\subset v.

Compare this result with exercise 24 of IV, p. 352.

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Machine translation of the English, by gpt-5.4. Not checked by a person.