Density Separability and Convolution in $L^p$ — Examples

For fixed $x$, the integrand is $1$ exactly when [L1, given, algebra] $y\in [0,1]\cap [x-1,x]$. Therefore $(f\ast f)(x)$ is the length of that overlap interval.

If $0\le x\le 1$, the overlap is $[0,x]$, so $(f\ast f)(x)=x$. If [step 1.1, algebra] $1\le x\le 2$, the overlap is $[x-1,1]$, so $(f\ast f)(x)=2-x$. For $x\le 0$ or $x\ge 2$, there is no overlap, so $(f\ast f)(x)=0$.

Since $\mathrm{supp}(f)=[0,1]$, [L2] predicts support inside [L2, step 2.1] $[0,1]+[0,1]=[0,2]$, exactly as the explicit computation shows.

The change of variables $u=(x-y)\mathrm{/}\epsilon$ gives the displayed formula. [L1, given, algebra] Because $\mathrm{supp}({\phi }_{\epsilon })\subseteq [-\epsilon ,\epsilon ]$, the convolution vanishes unless the interval $[0,1]$ meets $[x-\epsilon ,x+\epsilon ]$, namely unless $x\in [-\epsilon ,1+\epsilon ]$.

If $x\in [\epsilon ,1-\epsilon ]$, then the whole support of [L1, step 1.1, algebra] ${\phi }_{\epsilon }(x-\cdot )$ lies inside $[0,1]$, so $(f\ast {\phi }_{\epsilon })(x)=\int {\phi }_{\epsilon }=1.$ Near $x=0$ and $x=1$, only part of the kernel fits inside $[0,1]$, producing the two smooth transition layers.

By [L2], every $f\ast {\phi }_{\epsilon }$ is smooth; the cases [L2, L3, step 2.1] $\epsilon =1\mathrm{/}4$ and $\epsilon =1\mathrm{/}8$ differ only in the width of the two boundary layers, with the smaller $\epsilon$ giving the sharper transition.

Specialize [L1] to $n=1$ and $p=2$. The rational boxes in one dimension are [L1, given] the rational half-open intervals $(a,b]$, and restricting to those contained in $[0,1]$ still leaves a countable family.

Extend a function on $[0,1]$ by $0$ outside $[0,1]$. Then approximation in [step 1.1, algebra] $L^2(\mathbb{R})$ by rational interval step functions supported in $[0,1]$ restricts back to approximation in $L^2([0,1])$.

Therefore the stated family is an explicit countable dense subset of [step 2.1] $L^2([0,1])$.

Assume $s<t$. Then [L1, given, algebra] ${\mathbf{1}}_{[0,t]}-{\mathbf{1}}_{[0,s]}$ equals $1$ on $(s,t]$, a set of positive measure.

Hence [step 1.1, algebra]

The same conclusion holds when $t<s$ by symmetry.

So any two distinct members of the family are distance $1$ apart, which is [L1, step 2.1] exactly the one-separated property used in [L1].

The previous tent-function computation gives [L1, given, algebra] $(f\ast g)(x)=\{\begin{array}{ll}x,& 0\le x\le 1,\\ 2-x,& 1\le x\le 2,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$

Therefore [step 1.1, algebra] $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_2^2={\int }_0^1{x}^{2}dx+{\int }_1^2(2-x{)}^{2}dx=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\mathrm{.}$

Since $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_2=1$, this gives [L1, step 2.1] $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_2=\sqrt{2\mathrm{/}3}\le 1=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2$, exactly as [L1] predicts.

The change of variables $u=x\mathrm{/}\epsilon$ gives [L1, given, algebra] ${\int }_{{\mathbb{R}}^n}{G}_{\epsilon }(x)dx={\int }_{{\mathbb{R}}^n}(2\pi {)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2}du=1,$ so every kernel has mass one and $\mathrm{\parallel }{G}_{\epsilon }{\mathrm{\parallel }}_1=1$.

For every $\delta >0$, [step 1.1, algebra] ${\int }_{\mathrm{\mid }x\mathrm{\mid }>\delta }{G}_{\epsilon }(x)dx={\int }_{\mathrm{\mid }u\mathrm{\mid }>\delta \mathrm{/}\epsilon }(2\pi {)}^{-n\mathrm{/}2}{e}^{-\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2}du⟶0$ as $\epsilon \to 0^{+}$, because the integration region escapes to infinity against an integrable Gaussian tail.

Steps 1.1 and 2.1 verify the defining clauses of [L1], so the Gaussian [L1, step 1.1, step 2.1] family is an $L^1$ approximate identity even though it is not compactly supported.

The constant function $1$ belongs to $L^{\mathrm{\infty }}({\mathbb{R}}^n)$ but not to [L1] $C_0({\mathbb{R}}^n)$, because it does not vanish at infinity.

By [L1], every $L^{\mathrm{\infty }}$-limit of compactly supported continuous functions [L1, step 1.1] lies in $C_0({\mathbb{R}}^n)$. Therefore $1$ cannot lie in the closure of $C_c({\mathbb{R}}^n)$, and the claim is false.

For each $x\in X$, let $e_x:={\mathbf{1}}_{\{x\}}$. Since [L2, given, algebra] $\mu (\{x\})=1$, each $e_x$ lies in $L^p(\mu )$. If $x\mathrm{\ne }y$, then $\mathrm{\parallel }{e}_x-e_y{\mathrm{\parallel }}_p^p=\mathrm{\mid }1{\mathrm{\mid }}^p+\mathrm{\mid }-1{\mathrm{\mid }}^p=2,$ so $\mathrm{\parallel }{e}_x-e_y{\mathrm{\parallel }}_p=2^{1\mathrm{/}p}$.

Thus the uncountable family $\{e_x:x\in X\}$ is pairwise [L1, L2, step 1.1, algebra] $2^{1\mathrm{/}p}$-separated. No countable set can be dense in a metric space containing uncountably many disjoint balls of radius $2^{1\mathrm{/}p}\mathrm{/}3$. So this $L^p(\mu )$ is not separable, contradicting the claim.

Therefore the unrestricted statement is false; [L1] records the correct [L1, step 2.1] hypothesis ledger.

Let $f:={\mathbf{1}}_{[0,1]}$ on $\mathbb{R}$. For every $h\mathrm{\ne }0$, the [L1, given, algebra] functions $f$ and ${\tau }_{h}f$ differ by $1$ on a set of positive measure: for $0<h<1$, for instance, on $(1,1+h]$ one has ${\tau }_{h}f=1$ and $f=0$.

Therefore [step 1.1, algebra]

for every $h\mathrm{\ne }0$, so the norm does not tend to $0$ as $h\to 0$.

Hence translation is not continuous in $L^{\mathrm{\infty }}$, which is exactly why [L1, step 2.1] [L1] stops at $p<\mathrm{\infty }$.

The function $x\mapsto 1\mathrm{/}(\mathrm{\mid }x\mathrm{\mid }(\log (e\mathrm{/}\mathrm{\mid }x\mathrm{\mid }){)}^2)$ is integrable near $0$, so [L1, given, algebra] $f,g\in L^1(\mathbb{R})$.

At $x=0$ one has [step 1.1, algebra] ${\int }_{\mathbb{R}}\mathrm{\mid }f(-y)g(y)\mathrm{\mid }dy={\int }_{-1\mathrm{/}2}^{1\mathrm{/}2}\frac{dy}{\mathrm{\mid }y{\mathrm{\mid }}^2(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^4}\mathrm{.}$ The single point $y=0$ is irrelevant to Lebesgue integrability, so it is enough to inspect the punctured interval $(0,1\mathrm{/}2)$. There the substitution $u=\log (e\mathrm{/}y)$ gives $y=e^{1-u}$ and $dy=-e^{1-u}du$, hence ${\int }_0^{1\mathrm{/}2}\frac{dy}{y^2(\log (e\mathrm{/}y){)}^4}={\int }_{\log (2e)}^{\mathrm{\infty }}\frac{e^{u-1}}{u^4}du=\mathrm{\infty }\mathrm{.}$ Thus $(f\ast g)(0)$ is not defined as an absolutely convergent Lebesgue integral.

Therefore the convolution of two $L^1$ functions need not be defined at [L1, step 2.1] every point; [L1] correctly states only almost-everywhere existence.

Since [L1, given, algebra] ${\int }_{\mathbb{R}}(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-3\mathrm{/}2}dx<\mathrm{\infty },$ the functions $f$ and $g$ lie in $L^2(\mathbb{R})$.

For $x>2$, [step 1.1, algebra] $(f\ast g)(x)={\int }_{\mathbb{R}}(1+\mathrm{\mid }x-y\mathrm{\mid }{)}^{-3\mathrm{/}4}(1+\mathrm{\mid }y\mathrm{\mid }{)}^{-3\mathrm{/}4}dy\ge {\int }_1^{x-1}{x}^{-3\mathrm{/}4}{x}^{-3\mathrm{/}4}dy=\frac{x-2}{x^{3\mathrm{/}2}}\mathrm{.}$ So for large $x$, $(f\ast g)(x)\ge c{x}^{-1\mathrm{/}2}$ for some $c>0$.

But [L1, step 2.1] ${\int }_2^{\mathrm{\infty }}{x}^{-1}dx=\mathrm{\infty },$ so the lower bound from step 2.1 shows $f\ast g\mathrm{\notin }{L}^2(\mathbb{R})$. Hence the statement refuted above is false.

Fact [L2] gives a continuous map $T$ and a Lebesgue measurable function [L2] $u$ such that $u\circ T$ is not Lebesgue measurable. So the slogan "Lebesgue measurability survives every continuous change of variables" is false.

The displayed claim relies on exactly that false slogan. Even if some [L1, step 1.1] particular maps used in convolution behave better, the blanket justification for dropping the Borel-representative discipline recorded in [L1] fails.

Therefore the displayed claim is false.