The $L^p$ Spaces Holder Minkowski and Riesz Fischer: Examples and Counterexamples

Because $\mathrm{\mid }{f}_0{\mathrm{\mid }}^p=x^{-ap}{\chi }_{(0,1)}$ and $\mathrm{\mid }{f}_{\mathrm{\infty }}{\mathrm{\mid }}^p=x^{-ap}{\chi }_{(1,\mathrm{\infty })}$, [L1] reduces both claims to the improper integrals of $x^{-ap}$.

On $(0,1)$, the antiderivative is $x^{1-ap}\mathrm{/}(1-ap)$ when $ap\mathrm{\ne }1$, so the improper integral converges when $1-ap>0$, that is, when $ap<1$. If $ap=1$, it is ${\int }_0^{1}dx\mathrm{/}x$, which diverges by [L2]. If $ap>1$, then $x^{-ap}\ge x^{-1}$ on $(0,1)$, so divergence follows from [L2] and the comparison test [L3].

On $(1,\mathrm{\infty })$, the same antiderivative converges when $1-ap<0$, that is, when $ap>1$. If $ap=1$, it is again ${\int }_1^{\mathrm{\infty }}dx\mathrm{/}x$, which diverges by [L2]. If $ap<1$, then $x^{-ap}\ge x^{-1}$ for $x\ge 1$, so divergence follows from [L2] and the comparison test [L3].

Steps 2.1 and 2.2 prove the two threshold claims.

By [L1], the sequence $(a_k)$ lies in ${\mathrm{\ell }}^p$ exactly when [L1, given] ${\sum }_{k=0}^{\mathrm{\infty }}\mathrm{\mid }{a}_k{\mathrm{\mid }}^p={\sum }_{k=1}^{\mathrm{\infty }}{k}^{-ap}$ converges.

By [L2], that series converges exactly when $ap>1$.

Extending $(x_0,\dots ,x_{n-1})$ by zeros outside $\{0,\dots ,n-1\}$ turns it [L1, given] into a sequence in the setting of [L1]. The resulting ${\mathrm{\ell }}^p$ and $L^{\mathrm{\infty }}$ formulas are exactly the two displayed expressions.

In the ranges stated in [L2], those expressions are exactly the published [step 1.1, L2] norms on ${\mathbb{R}}^n$. For other real $p>0$, step 1.1 still gives the displayed $L^p$ functional, without claiming that the earlier finite-dimensional page called it a norm. ∎

The pointwise supremum is $1$ because $f(x)=1$ on every rational point of $[0,1]$.

If $0<\epsilon <1$, then $\{f>\epsilon \}=\mathbb{Q}\cap [0,1]$, which is null by [L2]; if $\epsilon \ge 1$, then $\{f>\epsilon \}=\mathrm{\varnothing }$, which is also null. Hence every $\epsilon >0$ is an essential bound in the sense of [L1], and therefore $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$.

Steps 1.1 and 1.2 prove the two claims.

The functions satisfy [given] $\mathrm{\mid }f{\mathrm{\mid }}^p=a^p{\chi }_{[0,1]},\mathrm{\mid }g{\mathrm{\mid }}^q=b^q{\chi }_{[0,1]},$ so $\mathrm{\mid }f{\mathrm{\mid }}^p=(a^p\mathrm{/}{b}^q)\mathrm{\mid }g{\mathrm{\mid }}^q$ almost everywhere.

Applying [L1] gives equality in Holder: [L1, step 1.1] ${\int }_0^1\mathrm{\mid }fg\mathrm{\mid }d\lambda =\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ ∎

Direct computation gives [given, algebra] $\mathrm{\parallel }f{\mathrm{\parallel }}_p^p={\int }_0^1\mathrm{\mid }f{\mathrm{\mid }}^{p}d\lambda =\frac{2^p}{2}+\frac{1^p}{2}=2^{p-1}+\frac{1}{2}\mathrm{.}$

Hence [L1, step 1.1, algebra] $\mathrm{\parallel }f{\mathrm{\parallel }}_p={(2^{p-1}+\frac{1}{2})}^{1\mathrm{/}p}=2{(\frac{1}{2}+2^{-p-1})}^{1\mathrm{/}p}\mathrm{.}$ For $p\ge 1$, its bracket lies between $1\mathrm{/}2$ and $1$, so its $1\mathrm{/}p$ power lies between $2^{-1\mathrm{/}p}$ and $1$ and therefore tends to $1$. Thus $\mathrm{\parallel }f{\mathrm{\parallel }}_p\to 2$. This agrees with [L1] because the essential supremum of $f$ is $2$. ∎

Since $f$ and $g$ are disjoint indicators of sets of measure $1\mathrm{/}2$, [given, algebra] $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=\frac{1}{2},\mathrm{\parallel }f+g{\mathrm{\parallel }}_2^2=1,\mathrm{\parallel }f-g{\mathrm{\parallel }}_2^2=1.$

Therefore [L1, step 1.1] $\mathrm{\parallel }f+g{\mathrm{\parallel }}_2^2+\mathrm{\parallel }f-g{\mathrm{\parallel }}_2^2=1+1=2=2\cdot \frac{1}{2}+2\cdot \frac{1}{2}=2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+2\mathrm{\parallel }g{\mathrm{\parallel }}_2^2,$ exactly as [L1] predicts. ∎

Choose $a=3\mathrm{/}4$ and set $f(x):=x^{-3\mathrm{/}4}{\chi }_{(0,1)}(x)$. Then [L1, given] $a\cdot 1=3\mathrm{/}4<1$, so [L1] gives $f\in L^1$. But $a\cdot 2=3\mathrm{/}2>1$, so [L1] also gives $f\mathrm{\notin }{L}^2$.

Thus $f$ is an $L^1$ function on $\mathbb{R}$ that does not lie in $L^2$, [step 1.1] refuting the claim. ∎

Choose $a=3\mathrm{/}4$ and set $f(x):=x^{-3\mathrm{/}4}{\chi }_{(1,\mathrm{\infty })}(x)$. Then [L1] $a\cdot 2=3\mathrm{/}2>1$, so [L1] gives $f\in L^2$. But $a\cdot 1=3\mathrm{/}4≯1$, so gives $f\mathrm{\notin }{L}^1$.

Thus $f$ is an $L^2$ function on $\mathbb{R}$ that does not lie in $L^1$, [step 1.1] refuting the claim. ∎

On $[0,1]$ with Lebesgue measure, let [L1, given, algebra] $f:={\chi }_{[0,1\mathrm{/}2]}$ and $g:={\chi }_{(1\mathrm{/}2,1]}$. Then $\mathrm{\parallel }f{\mathrm{\parallel }}_{1\mathrm{/}2}={\left({\int }_0^{1\mathrm{/}2}1d\lambda \right)}^2=\frac{1}{4},\mathrm{\parallel }g{\mathrm{\parallel }}_{1\mathrm{/}2}=\frac{1}{4},$ while $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{1\mathrm{/}2}={\left({\int }_0^{1}1d\lambda \right)}^2=1.$

Therefore [step 1.1, L1] $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{1\mathrm{/}2}=1>\frac{1}{4}+\frac{1}{4}=\mathrm{\parallel }f{\mathrm{\parallel }}_{1\mathrm{/}2}+\mathrm{\parallel }g{\mathrm{\parallel }}_{1\mathrm{/}2},$ so the triangle inequality fails. This is exactly the phenomenon summarized in [L1]. ∎

Let $f:={\chi }_{\mathbb{Q}\cap [0,1]}$. Then $f$ is not the zero function, [given] because $f(q)=1$ for every rational $q\in [0,1]$.

By [L1], the support of $f$ is null, so $f=0$ almost everywhere. Therefore [L1, L2] [L2] gives $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$ in every finite-$p$ range treated on the page, and also $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=0$.

Thus a nonzero measurable function can have zero $L^p$ seminorm.

Let $f_n:=h$ for every $n$, where $h$ is the function from [L1]. Then [L1, L2] $(f_n)$ is constant, hence Cauchy in the representative $p$-distance.

Also [L1, step 1.1] $\mathrm{\parallel }{f}_n-h{\mathrm{\parallel }}_p=0,\mathrm{\parallel }{f}_n-0{\mathrm{\parallel }}_p=\mathrm{\parallel }h{\mathrm{\parallel }}_p=0$ for every $n$. So the same sequence converges both to $h$ and to $0$, even though those two functions are distinct pointwise.

Therefore the representative-space distance does not have unique limits [step 2.1] before passing to almost-everywhere classes. ∎

Taking $p=1$ and $r=2$, [L1] already contradicts the claim.

The reverse failure [L2] shows why the line supports neither global [L2] inclusion.

Hence the displayed universal inclusion statement is false.

Let $f$ be the function from [L1]. Then $f\mathrm{\ne }0$ pointwise but [L1] $\mathrm{\parallel }f{\mathrm{\parallel }}_p=0$.

This violates the separation axiom in [L2], so the $p$-seminorm on [step 1.1, L2] ${\mathcal{L}}^p(\mu )$ is not a norm. ∎

The single function from [L1] already violates the claimed equality.

Therefore the statement is false.

The sequence from [L1] converges in $L^1$, hence is Cauchy in $L^1$.

The same source records representatives that fail to converge almost [L1, step 1.1] everywhere. So representatives of a Cauchy sequence in $L^p$ need not converge pointwise almost everywhere.

Therefore the universal claim is false.

The functions from [L1] violate the triangle inequality required by [L2].

Therefore $L^{1\mathrm{/}2}(\mu )$ with $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{1\mathrm{/}2}$ is not a normed space.

Let $A=[0,1\mathrm{/}2]$, let $f:={\chi }_A$, and let $g:={\chi }_{[0,1]}$. Then [L1, given, algebra] ${\int }_0^1\mathrm{\mid }fg\mathrm{\mid }d\lambda =\lambda (A)=\frac{1}{2},\mathrm{\parallel }f{\mathrm{\parallel }}_1=\frac{1}{2},\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}=1.$ So equality holds in Holder: ${\int }_0^1\mathrm{\mid }fg\mathrm{\mid }d\lambda =\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

There is no constant $c$ with $f=cg$ almost everywhere, because on [L2, step 1.1] $A$ one would need $c=1$ while on $(1\mathrm{/}2,1]$ one would need $c=0$. This does not contradict [L2], because [L2] does not cover the endpoint $p=1$.

Thus equality in Holder does not force the functions themselves to be [step 1.1, step 2.1] proportional almost everywhere. ∎