Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 22H (Analysis of Functions)

If $f=g$ almost everywhere, then $|f|=|g|$ almost everywhere, so the integral defining $\Vert [f]{\Vert }_1$ is independent of the representative. Homogeneity and the triangle inequality follow from the corresponding pointwise properties of absolute value. Moreover,so this is a norm on the Lebesgue space of almost-everywhere equivalence classes.

The Riesz-Fischer theorem states that $L^p$ is complete for $1\le p\le \infty$. Applied with $p=1$, it says that every $\Vert \cdot {\Vert }_1$-Cauchy sequence of these equivalence classes converges in ${\mathcal{L}}^1({\mathbb{R}}^n)$. Thus ${\mathcal{L}}^1({\mathbb{R}}^n)$ is a Banach space.

Consider the inclusion mapIts graph is closed. Indeed, suppose $x_k\to x$ in $X$ and $J{x}_k\to y$ in $L^1$. By hypothesis, some subsequence of $(x_k)$ converges almost everywhere to $x$. From its $L^1$ convergence, that subsequence has a further subsequence converging almost everywhere to $y$. Uniqueness of almost-everywhere limits gives $x=y$ as $L^1$ classes.

Both spaces are Banach, so the closed graph theorem makes $J$ bounded. Therefore some $C>0$ satisfies

No. TakeThis is a Banach space, and $X$-convergence implies $L^2$ convergence, which has an almost-everywhere convergent subsequence. Thus the hypotheses hold.

However, letThe power singularity integrability test gives $f\in L^1$ but $f\notin L^2$. Its truncations $f_m=\min (f,m)$ belong to $X$ and converge to $f$ in $L^1$ by dominated convergence. Hence $(f_m)$ is $L^1$-Cauchy in $X$ but has no $L^1$ limit belonging to $X$. Therefore $X$ need not be complete in the $L^1$ norm.