Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 5

The sequence $(f_n)$ converges by uniform convergence to $f$ on $E$ ifThe index $M$ must work simultaneously for every $x\in E$.

Given $\epsilon >0$, choose $m\ge N$ so large that $|f_m(x)-f(x)|<\epsilon /3$ for every $x\in E$. The continuity of this one fixed $f_m$ at $x_0$ gives $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\epsilon /3$. The triangle inequality then givesThus $f$ is continuous at $x_0$, with continuity understood relative to $E$. This is the pointwise form of the uniform limit theorem.

No. Let $E=\mathbb{R}$, $x_0=0$ andThe supremum norm satisfies $\Vert f_n-f{\Vert }_{\infty }=1/n\to 0$, so there is uniform convergence to the continuous function $f$. But every $f_n$ is discontinuous at zero: $f_n(0)=1/n$ whereas $f_n(x)=0$ for every $x\ne 0$. No tail consists of functions continuous at $x_0$.

At $x=0$, $f_n(x)=0$. For fixed $x>0$,Thus the pointwise limit is zero everywhere on $[0,1]$. However, with $t=nx$, the inequality $1+t^2\ge 2t$ gives $f_n(x)\le 1/2$, and equality occurs at $x=1/n\in [0,1]$. Henceso there is no uniform convergence. The moving evaluation point $1/n$ captures a peak missed by each fixed positive input.

The finite geometric series identity gives, for each $|x|<1$,For every fixed $n$, the absolute value of this error tends to infinity as $x\to 1^{-}$. Thusand there is no uniform convergence on $(-1,1)$.

Yes. By uniform convergence, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in \mathbb{R}$. ThenThe polynomial $p_n-p_N$ is bounded on the whole real line. The supplied growth property of nonconstant polynomials implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the limit of a sequence in $p_n(x)=p_N(x)+c_n$ to obtainThis is a polynomial. The uniform limits of polynomials on the real line therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.