Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 1 / c

Extend the coefficient lists of each polynomial by zeros. Then $\rho (p,q)={\sum }_{k\ge 0}|a_k-b_k|$ is finite because only finitely many terms are nonzero. It is nonnegative and symmetric. Moreover, $\rho (p,q)=0$ holds exactly when every coefficient agrees, which by uniqueness of polynomial coefficients is equivalent to $p=q$.

If $r(t)=\sum c_k{t}^k$, the triangle inequality for the absolute value givesThus $\rho$ satisfies all the metric axioms. It is the metric induced by the polynomial coefficient sum norm.

The space of polynomials is not complete for the supremum norm metric $d$. ConsiderThe geometric series has sum $F(t)=1/(1-t/2)$, and for $0\le t\le 1$ its tail satisfiesThus $(p_n)$ converges uniformly to $F$ and is a Cauchy sequence for $d$.

There is no polynomial $p$ agreeing with $F$ on $[0,1]$: otherwise the polynomial $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero polynomial. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the uniform limit, $(p_n)$ has no limit in $X$. This proves the claimed failure of completeness.

For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.