Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 12F (Topics In Analysis)

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

The Chebyshev alternation theorem says that a polynomial $P$ of degree at most $n$ is a best uniform approximation to $f\in C([0,1])$ if and only if there are $n+2$ pointsand a sign $\epsilon \in \{1,-1\}$ for whichThis is the equiripple criterion.

To prove sufficiency, put $E=\Vert f-P{\Vert }_{\infty }$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\Vert f-Q{\Vert }_{\infty }<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The intermediate value theorem then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The Lagrange root bound over a field forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\mathrm{deg}R\le n$. Therefore no $Q$ gives a smaller error, and $P$ is best.

As printed, this part contains an indexing error. The displayed product ends at $m-1$, but the hypotheses contain $m+1$ alternating intervals indexed by $0,\dots ,m$. On the first $m$ intervals the printed polynomial has sign $(-1{)}^i$, while on the last interval it has sign $(-1{)}^{m-1}$ rather than the required $(-1{)}^m$. For example, when $m=1$ the printed product is the constant $Q=1$, which cannot reduce a negative extremum of $g$ on the final interval. The intended polynomial isWe prove the stated conclusion with this correction.

Each root $w_j$ lies in the gap between the $(j-1)$st and $j$th active intervals. Therefore $Q$ has sign $(-1{)}^i$ throughout $[u_i,v_i]$. In particular, wherever $|g|=M$, the numbers $Q$ and $g$ have the same sign, and $Q$ is nonzero.

LetThese are compact sets. On their union, the continuous function $|Q|$ has a positive minimum. Hence, throughout some open neighbourhood $U$ of $K_{+}\cup K_{-}$, subtracting a sufficiently small positive multiple $\eta Q$ moves $g$ strictly towards zero and gives $|g-\eta Q|<M$. On the compact complement $[0,1]\setminus U$, continuity gives a uniform margin $|g|\le M-ϵ$ for some $ϵ>0$. Taking alsoshows that $|g-etaQ|<M$ there. Thusfor every sufficiently small $\eta >0$.

It remains to prove necessity in the Chebyshev alternation theorem. Let $P$ be a best approximation of degree at most $n$, put $g=f-P$, and let $M=\Vert g{\Vert }_{\infty }$. If $g$ has no sequence of $n+2$ extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into $m+1$ alternating compact groups with $m\le n$. Choose disjoint intervals $[u_i,v_i]$ containing those groups and separated by regions on which $|g|<M$. After replacing $g$ by $-g$ if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial $Q$ of degree $m\le n$ and a small $\eta >0$ such thatThis contradicts the assumed optimality of $P$. Therefore every best polynomial has the required $n+2$ alternating extrema.