Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 75 / 1

Assume $a<b$. A Chebyshev system of dimension $n+1$ is a family of real continuous functions $u_0,\dots ,u_n$ for which every linear combination $u={\sum }_{j=0}^n{c}_j{u}_j$ with coefficients not all zero has at most $n$ distinct zeros in the interval. In particular the functions are linearly independent: an identically zero nontrivial combination would violate that bound. Multiplicities are not counted in this definition.

For distinct points $x_0,\dots ,x_n$, form the evaluation matrix $A=(u_j(x_i){)}_{i,j=0}^n$. If its determinant vanished, a nonzero vector $c$ would satisfy $Ac=0$. Its associated combination would vanish at all $n+1$ points, contradicting the Chebyshev system property. Conversely, if the property failed, some nonzero coefficient vector would give a combination vanishing at $n+1$ distinct points. At those points $Ac=0$, so the evaluation matrix would be singular. We have proved the determinant criterion for a Chebyshev system:This is also the existence-and-uniqueness condition for interpolation from the span of the functions: any prescribed $n+1$ data values give a uniquely solvable linear system.

The exponents are real, as required for this family of real continuous functions. We prove the exponential Chebyshev system property by induction on the number of functions. A single nonzero multiple of an exponential function has no zero, so the base case holds.

Suppose the result holds for $n$ distinct real exponents, and consider a nontrivial combination $F(x)={\sum }_{j=0}^n{c}_j{e}^{{\lambda }_{j}x}$. If it had $n+1$ distinct zeros, so wouldsince the prefactor is everywhere positive. Applying Rolle's theorem between consecutive zeros gives at least $n$ distinct zeros ofThe $n$ exponents in this derivative are distinct. By induction, a nontrivial combination of them has at most $n-1$ zeros. Therefore every coefficient $c_j({\lambda }_j-{\lambda }_0)$ must be zero. Distinctness gives $c_j=0$ for $j\ge 1$. Then $g=c_0$ has a zero only if $c_0=0$, contradicting the original nontriviality. Consequently $F$ has at most $n$ zeros, and

Let $d$ be the dimension of a proposed real Chebyshev system on the circle, and regard its functions as continuous $2\pi$-periodic functions on the real line. Choose lifted points ${\theta }_0<{\theta }_1<\cdots <{\theta }_{d-1}<{\theta }_0+2\pi$ and set ${\theta }_d={\theta }_0+2\pi$. Move the points byFor every $s$ the points remain distinct on the circle. Indeed, each consecutive gap is a convex combination of positive gaps; the closing gap is also a convex combination of the original closing gap and ${\theta }_1-{\theta }_0$. The evaluation determinantis therefore nonzero for every $s$ by the determinant criterion for a Chebyshev system. It is continuous, so the intermediate value theorem forbids a sign change. At $s=1$, however, periodicity makes its rows the cyclic permutation of the initial rows, giving $D(1)=(-1{)}^{d-1}D(0)$. For even $d$ this reverses the sign, a contradiction. Hence every such periodic Chebyshev system has odd dimension; none has even dimension. This determinant argument avoids assuming differentiability or simple zeros of the functions.