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双対性議論(duality argument)について

  
   

p\in[1,\infty)に対して,p'=\bra{1-\f{1}{p}}^{-1}p'\in(1,\infty]を定める(すなわち,pp'はHölder共役).このとき,L^{p}の共役空間(双対空間)(L^{p})^{*}L^{p'}と同型であることはよく知られているが,この事実を示すことは簡単ではない.

しかし,この双対的な議論をするとき,(L^{p})^{*}\cong L^{p'}まで用いる必要はなく,次の事実だけで十分なことも多い:

任意のv\in L^{p'}(\Omega)に対して,

\|v\|_{L^{p'}(\Omega)} =\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx}

が成り立つ.

この事実は比較的容易に証明することができる.

なお,この記事では,\Omega\subset\R^{N}に対して,I_{\Omega}\Omega\subset\R^{N}上の定義関数とする.

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準備

まずは,「Hölder共役」を定義する.

[定義1] p\in[1,\infty]に対して,

\f{1}{p'}+\f{1}{p}=1 \Lra p'=\bra{1-\f{1}{p}}^{-1}

で定まるp'\in[1,\infty]pのHölder共役という.ただし,p=1のときはp'=\inftyとみなし,p=\inftyのときはp'=1とみなす.

次のL^{p}L^{q}空間を定義する.このL^{p}L^{q}空間は,例えば発展方程式で空間\R^Nに関してL_{x}^{p},時間\Rに関してL_{t}^{q}である場合などに用いる.

この定義は冒頭で述べた双対性の証明には用いないので,ひとまず飛ばして読んでも問題はない.

[定義2] i=1,2に対して,\Omega_{i}\subset\R^{N_{i}}を開集合,p_{i}\in[1,\infty]とする.このとき,L^{p_{1}}(\Omega_{1},L^{p_{2}}(\Omega_{2}))は,\Omega_{1}\times\Omega_{2}を定義域とし,\Omega_{1}上ほとんど至るところでL^{p_{2}}(\Omega_{2})に値をとるL^{p_{1}}空間である:

L^{p_{1}}(\Omega_{1},L^{p_{2}}(\Omega_{2})) :=\set{u:\Omega_{1}\times\Omega_{2}\to\C} {\|u\|_{L^{p_{1}}(\Omega_{1},L^{p_{2}}(\Omega_{2}))}<\infty}
\|u\|_{L^{p_{1}}(\Omega_{1},L^{p_{2}}(\Omega_{2}))} :=\bra{\dint_{\Omega_{1}}\|u(x_{1},\cdot)\|_{L^{p_{2}}(\Omega_{2})}^{p_{1}}\,dx_{1}}^{1/p_{1}}

Lp空間の双対性議論

冒頭で述べた双対性を証明する.

[定理3] \Omega\subset\R^Nを開集合とする.p\in[1,\infty)とし,p'\in(1,\infty]pのHölder共役とする.このとき,任意のv\in L^{p'}(\Omega)に対して,

\|v\|_{L^{p'}(\Omega)} =\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx}

が成り立つ.

[証明]

Hölderの不等式より,任意のv\in L^{p'}(\Omega)に対して,

\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx} \le\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\|u\|_{p}\|v\|_{p'} =\|v\|_{p'}

だから,あとは

\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx} \ge\|v\|_{p'}

を示せば題意が従う.

これをp=1のとき,p\in(1,\infty)のときに分けて示す.

Step.1

p=1のとき

p'=\inftyである.任意の\epsilon>0に対して,ある可測集合\Omega'\subset\Omega0<|\Omega'|<\inftyを満たすものが存在して,

x\in\Omega'\Ra|v(x)|>\|v\|_{\infty}-\epsilon

が成り立つ.このとき,w:\Omega\to\C

w(x):=\dfrac{\overline{v(x)}}{|\Omega'||v(x)|}I_{\Omega'}(x)

で定めると,

\dint_{\Omega}|w(x)|\,dx =\f{1}{|\Omega'|}\int_{\Omega'}\,dx=1

すなわち\|w\|_{p}=1だから,

\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx}
\ge\abs{\dint_{\Omega}w(x)v(x)\,dx}
=\abs{\f{1}{|\Omega'|}\dint_{\Omega'}\f{\overline{v(x)}v(x)}{|v(x)|}\,dx}
=\f{1}{|\Omega'|}\dint_{\Omega'}|v(x)|\,dx
>\f{1}{|\Omega'|}\dint_{\Omega'}(\|v\|_{\infty}-\epsilon)\,dx
=\|v\|_{\infty}-\epsilon

となって,\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx}\ge\|v\|_{\infty}が従う.

Step.2

p\in(1,\infty)のとき

p'\in(1,\infty)である.\theta_{x}:=\arg v(x)とし,w:\Omega\to\Cw(x):=\f{|v(x)|^{p'-1}}{\|v\|_{p'}^{p'/p}}e^{-i\theta_{x}}で定める.このとき,

\bra{\dint_{\Omega}|w(x)|^{p}\,dx}^{1/p}
=\f{1}{\|v\|_{p'}^{p'/p}}\bra{\dint_{\Omega}|v(x)|^{p(p'-1)}\,dx}^{1/p}
=\f{1}{\|v\|_{p'}^{p'/p}}\bra{\dint_{\Omega}|v(x)|^{p'}\,dx}^{1/p} =1

だから,\|w\|_{p}=1に注意すれば,

\sup\limits_{\|u\|_{L^{p}(\Omega)}=1}\abs{\dint_{\Omega}u(x)v(x)\,dx}
\ge\abs{\dint_{\Omega}w(x)v(x)\,dx}
=\abs{\f{1}{\|v\|_{p'}^{p'/p}}\dint_{\Omega}|v(x)|^{p'-1}e^{-i\theta_{x}}|v(x)|e^{i\theta_{x}}\,dx}
=\f{1}{\|v\|_{p'}^{p'/p}}\dint_{\Omega}|v(x)|^{p'}\,dx
=\|v\|_{p'}^{p'-\frac{p'}{p}}=\|v\|_{p'}

が従う.

[証明終]

LpLq空間の双対性議論

次に,L^{p}L^{q}空間に関する双対性を示す.

[定理4] i=1,2に対して,\Omega_{i}\subset\mathbb{R}^{N_{i}}を開集合とする.p_{i}\in[1,\infty)とし,p'_{i}\in(1,\infty]p_{i}のHölder共役とする.このとき,任意のv\in L^{p'_{1}}(\Omega_{1},L^{p'_{2}}(\Omega_{2}))に対して,

\|v\|_{L^{p'_{1}}(\Omega_{1},L^{p'_{2}}(\Omega_{2}))} =\sup\limits_{\|u\|_{L^{p_{1}}\bra{\Omega_{1},L^{p_{2}}(\Omega_{2})}}=1} \abs{\dint_{\Omega_{1}}\int_{\Omega_{2}} u(x_{1},x_{2})v(x_{1},x_{2})\,dx_{2}dx_{1}}

が成り立つ.

[証明]

x:=(x_{1},x_{2})X:=L^{p'_{1}}(\Omega_{1},L^{p'_{2}}(\Omega_{2}))Y:=L^{p_{1}}(\Omega_{1},L^{p_{2}}(\Omega_{2}))とする.Hölderの不等式より,任意のv\in Xに対して,

\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}}\bra{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}}
\le\sup\limits_{\|u\|_{Y}=1}\dint_{\Omega_{1}}\abs{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}dx_{1}
\le\sup\limits_{\|u\|_{Y}=1}\dint_{\Omega_{1}}\|u(x_{1},\cdot)\|_{p_{1}}\|v(x_{1},\cdot)\|_{p'_{1}}\,dx_{1}
\le\sup\limits_{\|u\|_{Y}=1}\|u\|_{Y}\|v\|_{X}
=\|v\|_{X}

だから,あとは

\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}}\bra{\int_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}} \ge\|v\|_{X}

を示せば題意が従う.

これをp_{1}=p_{2}=1のとき,p_{1}=1,p_{2}\in(1,\infty)のとき,p_{1},p_{2}\in(1,\infty)のときに分けて証明する.

Step.1

p_{1}=p_{2}=1のときを示す.

p'_{1}=p'_{2}=\inftyである.任意の\epsilon>0i\in\{1,2\}に対して,ある可測集合\Omega'_{i}\subset\Omega0<|\Omega'_{i}|<\inftyを満たすものが存在して,

x\in\Omega'_{1}\times\Omega_{2} \Ra |v(x)|>\|v(x_{1},\cdot)\|_{L^{\infty}(\Omega_{2})}-\epsilon >\|v\|_{X}-2\epsilon

が成り立つ.このとき,w:\Omega_{1}\times\Omega_{2}\to\Cw(x):=\f{\overline{v(x)}}{|\Omega'_{1}||\Omega'_{2}|v(x)}I_{\Omega'_{1}\times\Omega'_{2}}(x)で定めると,

\dint_{\Omega_{1}}\bra{\dint_{\Omega_{2}}|w(x)|\,dx_{2}}\,dx_{1}
=\f{1}{|\Omega'_{1}||\Omega'_{2}|}\dint_{\Omega'_{1}}\bra{\int_{\Omega'_{2}}\,dx_{2}}\,dx_{1}
=1

すなわち\|w\|_{Y}=1だから,

\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}}
\ge\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}w(x)v(x)\,dx_{2}}\,dx_{1}}
=\abs{\f{1}{|\Omega'_{1}||\Omega'_{2}|}\dint_{\Omega'_{1}} \bra{\dint_{\Omega'_{2}}\f{\overline{v(x)}v(x)}{v(x)}\,dx_{2}}\,dx_{1}}
=\f{1}{|\Omega'_{1}||\Omega'_{2}|}\dint_{\Omega'_{1}} \bra{\dint_{\Omega'_{2}}|v(x)|\,dx_{2}}\,dx_{1}
>\f{1}{|\Omega'_{1}||\Omega'_{2}|}\dint_{\Omega'_{1}} \bra{\dint_{\Omega'_{2}}(\|v\|_{X}-\epsilon)\,dx_{2}}\,dx_{1}
=\|v\|_{X}-2\epsilon

だから,\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}}\ge\|v\|_{X}が従う.

Step.2

p_{1}=1,p_{2}\in(1,\infty)のときを示す.

p'_{1}=\infty,p'_{2}\in(1,\infty)である.任意の\epsilon>0に対して,ある可測集合\Omega'_{1}\subset\Omega_{1}0<|\Omega'_{1}|<\inftyを満たすものが存在して,

x\in\Omega'_{1}\times\Omega_{2} \Ra \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})} >\|v\|_{X}-\epsilon

が成り立つ.このとき,\theta_{x}:=\arg v(x)とし,w:\Omega_{1}\times\Omega_{2}\to\C

w(x):=\f{|v(x)|^{p'_{2}-1}}{|\Omega'_{1}| \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}/p_{2}}} e^{-i\theta_{x}}I_{\Omega'_{1}\times\Omega_{2}}(x)

で定めると,

\dint_{\Omega_{1}}\bra{\dint_{\Omega_{2}} |w(x)|^{p_{2}}\,dx_{2}}^{1/p_{2}}\,dx_{1}
=\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}/p_{2}}} \bra{\dint_{\Omega_{2}}|v(x_{1},x_{2})|^{p_{2}(p'_{2}-1)}\,dx_{2}}^{1/p_{2}}\,dx_{1}
=\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}/p_{2}}} \bra{\dint_{\Omega_{2}}|v(x_{1},x_{2})|^{p'_{2}}\,dx_{2}}^{1/p_{2}}\,dx_{1}
=\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}}\,dx_{1}=1

すなわち\|w\|_{Y}=1だから,

\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}} \ge\abs{\dint_{\Omega_{1}}\bra{\dint_{\Omega_{2}}w(x)v(x)\,dx_{2}}\,dx_{1}}
=\abs{\dfrac{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \bra{\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}/p_{2}}} \dint_{\Omega_{2}}|v(x_{1},x_{2})|^{p'_{2}-1}e^{-i\theta_{x}}v(x)\,dx_{2}}\,dx_{1}}
=\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \bra{\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}/p_{2}}} \dint_{\Omega_{2}}|v(x_{1},x_{2})|^{p'_{2}}\,dx_{2}}\,dx_{1}
=\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}-\frac{p'_{2}}{p_{2}}}\,dx_{1} =\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}\,dx_{1}
>\f{1}{|\Omega'_{1}|}\dint_{\Omega'_{1}} (\|v\|_{X}-\epsilon)\,dx_{1} =\|v\|_{X}-\epsilon

だから,\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}}\bra{\int_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}}\ge\|v\|_{X}が従う.

Step.3

p_{1},p_{2}\in(1,\infty)のときを示す.

p'_{1},p'_{2}\in(1,\infty)である.\theta_{x}:=\arg v(x)とし,w:\Omega_{1}\times\Omega_{2}\to\mathbb{C}

w(x) :=\f{|v(x)|^{p'_{2}-1}}{\|v\|_{X}^{p'_{1}/p_{1}} \|v(x_{1},\cdot)\|_{L^{p'_{2}} (\Omega_{2})}^{1-p'_{1}+p'_{2}/p_{2}}}e^{-i\theta_{x}}

で定めると,

\bra{\dint_{\Omega_{1}}\abs{\dint_{\Omega_{2}} |w(x)|^{p_{2}}\,dx_{2}}^{p_{1}/p_{2}}\,dx_{1}}^{1/p_{2}}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}} \bra{\dint_{\Omega'_{1}} \dfrac{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p_{1}(1-p'_{1}+p'_{2}/p_{2})}} \bra{\dint_{\Omega_{2}}|v(x)|^{p_{2}(p'_{2}-1)}\,dx_{2}}^{p_{1}/p_{2}}\,dx_{1}}^{1/p_{1}}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\bra{\dint_{\Omega'_{1}}\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}} (\Omega_{2})}^{p_{1}(1-p'_{1}+p'_{2}/p_{2})}} \bra{\int_{\Omega_{2}}|v(x)|^{p'_{2}}\,dx_{2}}^{p_{1}/p_{2}}\,dx_{1}}^{1/p_{1}}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}} \bra{\dint_{\Omega'_{1}}\|v(x_{1},\cdot)\|_{L^{p'_{2}} (\Omega_{2})}^{p_{1}(p'_{1}-1)}\,dx_{1}}^{1/p_{1}}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}} \bra{\dint_{\Omega'_{1}}\|v(x_{1},\cdot)\|_{L^{p'_{2}} (\Omega_{2})}^{p'_{1}}\,dx_{1}}^{1/p_{1}} =1

すなわち\|w\|_{Y}=1だから,

\sup\limits_{\|u\|_{Y}=1}\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}u(x)v(x)\,dx_{2}}\,dx_{1}}
\ge\abs{\dint_{\Omega_{1}} \bra{\dint_{\Omega_{2}}w(x)v(x)\,dx_{2}}\,dx_{1}}
=\abs{\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\dint_{\Omega_{1}} \bra{\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{1-p'_{1}+p'_{2}/p_{2}}} \dint_{\Omega_{2}}|v(x)|^{p'_{2}-1}e^{-i\theta_{x}}v(x)\,dx_{2}}\,dx_{1}}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\dint_{\Omega_{1}} \bra{\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{1-p'_{1}+p'_{2}/p_{2}}} \dint_{\Omega_{2}}|v(x)|^{p'_{2}}\,dx_{2}}\,dx_{1}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\dint_{\Omega_{1}} \bra{\f{1}{\|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{1-p'_{1}+p'_{2}/p_{2}}} \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{2}}}\,dx_{1}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\dint_{\Omega_{1}} \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{1}-1+p'_{2}(1-1/p_{2})}\,dx_{1}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\dint_{\Omega_{1}} \|v(x_{1},\cdot)\|_{L^{p'_{2}}(\Omega_{2})}^{p'_{1}}\,dx_{1}
=\f{1}{\|v\|_{X}^{p'_{1}/p_{1}}}\|v\|_{X}^{p'_{1}} =\|v\|_{X}^{p'_{1}(1-1/p_{1})} =\|v\|_{X}

が従う.

 [証明終]

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