Clarkson inequalities in both exponent ranges

Suppose $p\ge 2$ and let $r=p\mathrm{/}2\ge 1$. For $u,v\ge 0$, first $u^r+v^r\le (u+v{)}^r$: if $u+v>0$, divide by $(u+v{)}^r$ and use $0\le t\le 1\Rightarrow t^r\le t$ for $t=u\mathrm{/}(u+v)$ and $1-t=v\mathrm{/}(u+v)$; the zero case is equality. Also $(u+v{)}^r\le 2^{r-1}(u^r+v^r)$. For $r=1$ this is equality. For $r>1$, apply [F5] with conjugate $r$ and $r\mathrm{/}(r-1)$ to $(u,v)$ and $(1,1)$, then raise to the $r$th power.

Suppose $1<p<2$ and put $q=p\mathrm{/}(p-1)>2$. For $0\le t\le 1$, define $\mathrm{\Phi }(t)=\frac{((1+t{)}^q+(1-t{)}^q{)}^{1\mathrm{/}q}}{(1+t^p{)}^{1\mathrm{/}p}}\mathrm{.}$ We will prove $\mathrm{\Phi }(t)\le 2^{1\mathrm{/}q}$. Set $\alpha =p-1$ and $\beta =1\mathrm{/}(q-1)$; by [F1], $0<\alpha =\beta <1$. For $0<t<1$, put $G_1(t)=1-t^{2\alpha }-\alpha \beta t^{\alpha -1}+\alpha \beta t^{\alpha +1}$ and $H(t)=(\alpha +1)\beta t^2-2t^{\alpha +1}+(1-\alpha )\beta \mathrm{.}$ Direct differentiation gives G_1'(t)=\alpha t^{\alpha-2}H(t),\qquad H'(t)=2(\alpha+1)t(\beta-t^{\alpha-1})<0. \tag{2} because $t^{\alpha -1}>1>\beta$. The estimate $H(t)>0$ holds whenever $2t^{\alpha +1}<(1-\alpha )\beta$, while $H(1)=2(\beta -1)<0$. Continuity, [F4], and strict decrease therefore give a unique $t_0\in (0,1)$ at which $H$ vanishes. Thus $G_1$ increases before $t_0$ and decreases after it.

Suppose $1<p<2$, put $q=p\mathrm{/}(p-1)$, $r=q\mathrm{/}p>1$, and $s=r\mathrm{/}(r-1)$. For measurable representatives of $f,g$, let $A=\mathrm{\mid }f+g{\mathrm{\mid }}^p,B=\mathrm{\mid }f-g{\mathrm{\mid }}^p,U={\int }_{S}Ad\mu ,V={\int }_{S}Bd\mu ,$ and $M=(U^r+V^r{)}^{1\mathrm{/}r}$. These numbers are finite by [F7]. If $M=0$, the inequality below is immediate. If $M>0$, set $\lambda =(U\mathrm{/}M{)}^{r-1},\eta =(V\mathrm{/}M{)}^{r-1}\mathrm{.}$ Then ${\lambda }^s+{\eta }^s=1$ and $M=\lambda U+\eta V$. Two-term Hölder [F5], applied pointwise to $(\lambda ,\eta )$ and $(A,B)$, gives M=\int_S(\lambda A+\eta B)\,d\mu\le\int_S(A^r+B^r)^{1/r}\,d\mu. \tag{7}

For real or complex scalars $a,b$, [F6] and coordinate expansion give the scalar parallelogram identity $\mathrm{\mid }a+b{\mathrm{\mid }}^2+\mathrm{\mid }a-b{\mathrm{\mid }}^2=2(\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2)\mathrm{.}$ Apply both inequalities of step 1.1, first to $u=\mathrm{\mid }a+b{\mathrm{\mid }}^2,v=\mathrm{\mid }a-b{\mathrm{\mid }}^2$ and then to $u=\mathrm{\mid }a{\mathrm{\mid }}^2,v=\mathrm{\mid }b{\mathrm{\mid }}^2$. Since $2r=p$, this yields |a+b|^p+|a-b|^p\le2^{p-1}(|a|^p+|b|^p). \tag{1}

For every sufficiently small $t>0$, $G_1(t)\le 1+\alpha \beta -\alpha \beta t^{\alpha -1}<0;$ for example the final inequality holds whenever $t^{1-\alpha }<\alpha \beta \mathrm{/}(1+\alpha \beta )$. On the other hand $G_1(1)=0$, and strict decrease on $(t_0,1)$ gives $G_1(t)>0$ there. Consequently continuity and the monotonicity from step 1.2 give a unique $t_1\in (0,t_0)$ such that $G_1<0$ on $(0,t_1)$ and $G_1>0$ on $(t_1,1)$.

Define, for $0<t<1$, $G_2(t)=\log (1+t)+\beta \log (1-t^{\alpha })-\log (1-t)-\beta \log (1+t^{\alpha })\mathrm{.}$ Using [F2]–[F4] and simplifying over the positive common denominator gives G_2'(t)=\frac{2G_1(t)}{(1-t^2)(1-t^{2\alpha})}. \tag{3} For every $\epsilon >0$, $0<t<{\epsilon }^{1\mathrm{/}\alpha }$ implies $0<t^{\alpha }<\epsilon$; hence $t^{\alpha }\to 0$ as $t\downarrow 0$, and logarithm continuity proves that $G_2$ extends continuously by $G_2(0)=0$. By step 2.2 it first strictly decreases and then strictly increases. It eventually becomes positive: the derivative test applied to $1-t^{\alpha }-\alpha (1-t)$ gives $1-t^{\alpha }\ge \alpha (1-t)$, while $1+t^{\alpha }\le 2$; hence $\frac{1+t}{1-t}{\left(\frac{1-t^{\alpha }}{1+t^{\alpha }}\right)}^{\beta }\ge {\left(\frac{\alpha }{2}\right)}^{\beta }(1-t{)}^{\beta -1}>1$ whenever $0<1-t<(\alpha \mathrm{/}2{)}^{\beta \mathrm{/}(1-\beta )}$. By the logarithm and real-power laws, the logarithm of the left side is $G_2(t)$, so it is positive there. Thus [F4] gives a unique $t_2\in (t_1,1)$ such that $G_2<0$ on $(0,t_2)$ and $G_2>0$ on $(t_2,1)$.

Choose measurable representatives of $f$ and $g$; every pointwise inequality below is unchanged by modifying them on a null set. If $p\ge 2$, integrate (1), use [F7]–[F8], and divide by $2^p$: ${\parallel \frac{f+g}{2}\parallel }_p^p+{\parallel \frac{f-g}{2}\parallel }_p^p\le \frac{\mathrm{\parallel }f{\mathrm{\parallel }}_p^p+\mathrm{\parallel }g{\mathrm{\parallel }}_p^p}{2}\mathrm{.}$ The integrals are finite because $f\pm g\in L^p$ by the normed-space structure in [F7].

Put $G_3(t)=(1+t{)}^{q-1}(1-t^{p-1})-(1-t{)}^{q-1}(1+t^{p-1})\mathrm{.}$ The two terms are positive. Since $\log$ is strictly increasing, the sign of $G_3$ is the sign of the logarithm of their ratio, namely $(q-1)\log (1+t)+\log (1-t^{p-1})-(q-1)\log (1-t)-\log (1+t^{p-1})=(q-1)G_2(t)\mathrm{.}$ Another direct differentiation gives \Phi'(t)=\frac{\big((1+t)^q+(1-t)^q\big)^{1/q-1}}{(1+t^p)^{1+1/p}}G_3(t). \tag{4} The prefactor is positive. Step 3.1 therefore shows that $\mathrm{\Phi }$ decreases and then increases, so its maximum on $[0,1]$ is at an endpoint. Finally $\mathrm{\Phi }(0)=2^{1\mathrm{/}q},\mathrm{\Phi }(1)=\frac{2}{2^{1\mathrm{/}p}}=2^{1-1\mathrm{/}p}=2^{1\mathrm{/}q}\mathrm{.}$ This proves \big((1+t)^q+(1-t)^q\big)^{1/q}\le2^{1/q}(1+t^p)^{1/p}. \tag{5}

For arbitrary real $a,b$, the unordered pair $\{\mathrm{\mid }a+b\mathrm{\mid },\mathrm{\mid }a-b\mathrm{\mid }\}$ equals $\{\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid },\mathrm{\mid }\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }\}$. After swapping the two moduli and, when the larger one is nonzero, dividing by it, (5) gives (|a+b|^q+|a-b|^q)^{1/q}\le2^{1/q}(|a|^p+|b|^p)^{1/p}. \tag{6} The case $a=b=0$ is immediate.

The same estimate holds for complex $a,b$. The case where either is zero follows directly, so swap them if necessary and suppose $\mathrm{\mid }a\mathrm{\mid }\ge \mathrm{\mid }b\mathrm{\mid }>0$. Put $w=b\mathrm{/}a$, $r=\mathrm{\mid }w\mathrm{\mid }\le 1$, and $x=\mathrm{\mid }\mathrm{Re}w\mathrm{\mid }\le r$; the last inequality follows from $r^2=(\mathrm{Re}w{)}^2+(\mathrm{Im}w{)}^2$. The two terms below swap if $\mathrm{Re}w$ changes sign, so $\mathrm{\mid }1+w{\mathrm{\mid }}^q+\mathrm{\mid }1-w{\mathrm{\mid }}^q=(1+r^2+2x{)}^{q\mathrm{/}2}+(1+r^2-2x{)}^{q\mathrm{/}2}\mathrm{.}$ For $0\le c\le 1$, let $K(c)=(1+r^2+2rc{)}^{q\mathrm{/}2}+(1+r^2-2rc{)}^{q\mathrm{/}2}\mathrm{.}$ Its derivative is $K^{\mathrm{\prime }}(c)=qr((1+r^2+2rc{)}^{q\mathrm{/}2-1}-(1+r^2-2rc{)}^{q\mathrm{/}2-1})\ge 0.$ Thus $K(x\mathrm{/}r)\le K(1)=(1+r{)}^q+(1-r{)}^q$. Apply (5) to $r$, then multiply by $\mathrm{\mid }a\mathrm{\mid }$ using [F6], to obtain (6) over $\mathbb{C}$ as well.

Raise the complex-or-real scalar estimate (6) to the $p$th power. Since $pr=q$, it says pointwise that $(A^r+B^r{)}^{1\mathrm{/}r}\le 2^{p\mathrm{/}q}(\mathrm{\mid }f{\mathrm{\mid }}^p+\mathrm{\mid }g{\mathrm{\mid }}^p)\mathrm{.}$ Use this in (7), then use [F7]–[F8]: $(\mathrm{\parallel }f+g{\mathrm{\parallel }}_p^q+\mathrm{\parallel }f-g{\mathrm{\parallel }}_p^q{)}^{p\mathrm{/}q}\le 2^{p\mathrm{/}q}(\mathrm{\parallel }f{\mathrm{\parallel }}_p^p+\mathrm{\parallel }g{\mathrm{\parallel }}_p^p)\mathrm{.}$ Raising to $q\mathrm{/}p$ gives a factor $2$. Dividing by $2^q$ and using $1-q=-q\mathrm{/}p$ yields ${\parallel \frac{f+g}{2}\parallel }_p^q+{\parallel \frac{f-g}{2}\parallel }_p^q\le {\left(\frac{\mathrm{\parallel }f{\mathrm{\parallel }}_p^p+\mathrm{\parallel }g{\mathrm{\parallel }}_p^p}{2}\right)}^{q\mathrm{/}p}\mathrm{.}$

Step 3.2 proves the first claim, and step 6.1 proves the second for $1<p<2$. At $p=2$ one has $q=2$, and the scalar parallelogram identity in step 2.1 integrates to equality, so the overlapping endpoint belongs to both claims and the two displayed formulas coincide. The empty measure space, zero measure, and zero functions cause no exception: all their norms and all terms above are zero.