Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 7

Assume, as required for the displayed integrals, that the $f_n$ are Riemann integrable on $[a,b]$. Their uniform limit $f$ is also Riemann integrable: if $\sup |f-f_N|<\epsilon$, the upper and lower Darboux sums for a partition of an interval differ for $f$ by at most the corresponding difference for $f_N$ plus $2\epsilon (b-a)$. Choosing $N$ and then a sufficiently fine partition of an interval proves the Riemann integrability criterion for $f$.

The integral estimate then givesThis proves that uniform convergence preserves integrals on a compact interval. Some integrability hypothesis is necessary: uniform convergence alone cannot make the integral of an arbitrary starting function exist.

Let $f_n$ be the continuous triangular bumpand set $f_1=0$. At $x=0$ every term is zero, and each fixed $x>0$ eventually lies outside the support, so $f_n\to 0$ pointwise. The triangle has base $2/n$ and height $n$, henceThis concentrating triangular bump shows that pointwise convergence does not preserve integrals, even for continuous functions on a compact interval.

Take $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and the pointwise limit is $f(x)=0$ for $0\le x<1$, while $f(1)=1$. NeverthelessThe limiting function is Riemann integrable, since changing a value at one point does not change its Riemann integral. There is no uniform convergence: for every $n$, ${\sup }_{[0,1]}|f_n-f|=1$, with the supremum approached as $x\uparrow 1$ from below. Equivalently, the uniform limit theorem would require the limit to be continuous.

No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.