Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 2

First let $1\le p<\infty$, and take a Cauchy sequence $f_j$ in the Lp space, with functions identified almost everywhere. Choose a subsequence $f_{j_k}$ with $\Vert f_{j_{k+1}}-f_{j_k}{\Vert }_p\le 2^{-k}$ and put $a_k=f_{j_{k+1}}-f_{j_k}$. The triangle inequality givesBy the monotone convergence theorem, $A={\sum }_{k\ge 1}|a_k|$ belongs to $L^p$, and is finite almost everywhere. The series $f_{j_1}+{\sum }_{k\ge 1}{a}_k$ therefore converges almost everywhere to an $L^p$ function $f$. Applying the same argument to each absolute tail yieldsThe original sequence is Cauchy, so the triangle inequality extends convergence from this subsequence to the whole sequence.

For $p=\infty$, choose the same subsequence in essential supremum norm. Outside one null set, all its difference bounds hold simultaneously:The series then converges uniformly on that full-measure set to a bounded measurable $f$, and $\Vert f-f_{j_k}{\Vert }_{\infty }\le {\sum }_{\ell \ge k}{2}^{-\ell }$. Again the full sequence converges. Thus every $L^p$, $1\le p\le \infty$, is a Banach space, proving completeness of Lp spaces.

If either upper norm bound is nonpositive, there are no admissible functions with $\Vert f{\Vert }_q\ge C_q>0$, and the assertion is vacuous. For positive $C_p,C_r$, chooseSplitting the integral givesFor finite $r$, Holder inequality bounds the last term by $\Vert f{\Vert }_r^q\mu (A{)}^{1-q/r}\le C_r^q\mu (A{)}^{1-q/r}$. HenceFor $r=\infty$, the corresponding bound is ${\int }_A|f{|}^q\le C_r^q\mu (A)$. Thus an explicit choice proving the strict inequality isThis is a nonvanishing level-set bound between three Lp norms: the lower-$p$ bound prevents diffuse low-amplitude mass, and the higher-$r$ bound prevents arbitrarily concentrated spikes.

Work on $\mathbb{R}$ with Lebesgue measure. Without a uniform $L^p$ bound, useIt has $\Vert f_n{\Vert }_q=1$ and $\Vert f_n{\Vert }_r=n^{q/r-1}\le 1$, with $1/r=0$ for $r=\infty$. Each function still belongs to $L^p\cap L^r$, but its $L^p$ norm is unbounded. For every fixed $\epsilon >0$, its level set above $\epsilon$ is eventually empty, defeating any positive uniform $M$.

Without a uniform $L^r$ bound, use insteadNow $\Vert g_n{\Vert }_q=1$ and $\Vert g_n{\Vert }_p=n^{1-q/p}\le 1$, but for any fixed $\epsilon >0$ the high-amplitude level set has measure $n^{-q}\to 0$. These give the two distinct failures of the nonvanishing level-set bound between three Lp norms when either control is removed.