$l^p$为内积空间则$p=2$

hbghlyj

序列空间$l^p$为内积空间，那么 $p=2$。

证明 [搬运$l^p$ space not having inner product]
设$\mathbf{x} = (1,0,0,0,\ldots),\mathbf{y} = (0,1,0,0,\ldots)$. 则$\mathbf{x+y} = (1,1,0,0,\ldots),\mathbf{x-y} = (1,-1,0,0,\ldots)$. $$\|\mathbf{x}+\mathbf{y}\|_p^2 + \|\mathbf{x}-\mathbf{y}\|_p^2 = 2\times(1+1)^\frac{2}{p} = 2^{1+\frac{2}{p}}$$ 则 $$2(\|\mathbf{x}\|_p^2+\|\mathbf{y}\|_p^2) = 2\times(1+1) = 4$$ 若平行四边形恒等式成立, $$2^{1+\frac{2}{p}} = 4 \implies 1+\frac{2}{p} = 2 \implies p = 2$$

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相关问题: $L^p$空间为内积空间，那么$p=2$。

证明 [搬运How to prove that $L^p [0,1]$ isn't induced by an inner product? for $p\neq 2$]

For the case in which $1 \leq p < \infty$: Let $\displaystyle f = \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A$ and $\displaystyle g = \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B$ where $A, B$ both have nonzero finite measure, are disjoint and $\chi$ is the indicator function. Notice that $$\|f\|_p = \bigg( \int \vert f \vert^p \, d\mu \bigg)^{1/p} = \bigg( \int \frac{1}{\mu(A)} \chi_A \, d\mu \bigg)^{1/p} = \bigg( \int_A \frac{1}{\mu(A)} \, d\mu \bigg)^{1/p} = 1$$ and similarly $\|g\|_p = 1$. Observe that\begin{align*} \|f + g\|_p&= \bigg( \int \vert f + g \vert^p \, d\mu \bigg)^{1/p} \\ &=\bigg( \int \bigg \vert \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A + \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B \bigg \vert^p \, d\mu \bigg)^{1/p} \\ &= \bigg( \int \frac{1}{\mu(A)} \chi_A + \frac{1}{\mu(B)} \chi_B \, d\mu \bigg)^{1/p} \\ &= \bigg( \int_A \frac{1}{\mu(A)} \, d\mu + \int_B \frac{1}{\mu(B)} \, d\mu \bigg)^{1/p} \\ &= 2^{1/p} \end{align*}Note: We use the fact that $A$ and $B$ are disjoint and get $\chi_A \cdot \chi_B = 0$ when moving from line 2 to line 3 above. Similar reasoning shows $\|f - g\|_p = 2^{1/p}$. Recall that in an inner product space, the Parallelogram law holds: $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \|f\|_p^2 + 2 \|g\|_p^2$. But $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \cdot 2^{2/p}$ and $2 \|f\|_p^2 + 2 \|g\|_p^2 = 4$ which means that we must have $2^{2/p} = 2$ which only happens if $p = 2$. For the case in which $p = \infty$: Let $f = \chi_A$ and $g = \chi_B$ where $A, B$ are disjoint and both have nonzero measure. Notice that $\|f\|_\infty = \|g\|_\infty = 1$ and $\|f + g\|_\infty = \|f - g\|_\infty = 1$. The Parallelogram Law doesn't hold because $\|f + g\|_\infty^2 + \|f - g\|_\infty^2 = 2$ but $ 2 \|f\|_\infty^2 + 2 \|g\|_\infty^2 = 4$

Czhang271828

Hanner 不等式(以下 $f,g\in L^p(E)$, $1\leq p<\infty$)
$$
\begin{align*}
&\Vert f+g\Vert_p^p+\Vert f-g\Vert_p^p\leq \vert\Vert f\Vert_p+\Vert g\Vert_p\vert^p +\vert\Vert f\Vert_p-\Vert g\Vert_p\vert^p&&(p\geq 2),\\
&\Vert f+g\Vert_p^p+\Vert f-g\Vert_p^p\geq \vert\Vert f\Vert_p+\Vert g\Vert_p\vert^p +\vert\Vert f\Vert_p-\Vert g\Vert_p\vert^p&&(p\leq 2).
\end{align*}
$$
或等价地,
$$
\begin{align*}
&\dfrac{\vert\Vert f+g\Vert_p+\Vert f-g\Vert_p\vert^p+\vert\Vert f+g\Vert_p-\Vert f-g\Vert_p\vert^p}{2^p}\geq \Vert f\Vert_p^p+\Vert g\Vert_p^p&&(p\geq 2), \\
&\dfrac{\vert\Vert f+g\Vert_p+\Vert f-g\Vert_p\vert^p+\vert\Vert f+g\Vert_p-\Vert f-g\Vert_p\vert^p}{2^p}\leq \Vert f\Vert_p^p+\Vert g\Vert_p^p&&(p\leq 2), \\
\end{align*}
$$
Clarkson 不等式(以下 $f,g\in L^p(E)$, $1< p<\infty$, $\dfrac{1}{p}+\dfrac{1}{q}=1$)
$$
\begin{align*}
&\left\Vert\dfrac{f+g}{2}\right\Vert_p^p+\left\Vert\dfrac{f-g}{2}\right\Vert_p^p\leq\dfrac{\Vert f\Vert_p^p+\Vert g\Vert_p^p}{2}&&(p\geq 2),\\
&\left\Vert\dfrac{f+g}{2}\right\Vert_p^p+\left\Vert\dfrac{f-g}{2}\right\Vert_p^p\geq\dfrac{\Vert f\Vert_p^p+\Vert g\Vert_p^p}{2}&&(p\leq 2),\\
&\left\Vert\dfrac{f+g}{2}\right\Vert_p^q+\left\Vert\dfrac{f-g}{2}\right\Vert_p^q\geq\left(\dfrac{\Vert f\Vert_p^p+\Vert g\Vert_p^p}{2}\right)^{q-1}&&(p\geq 2),\\
&\left\Vert\dfrac{f+g}{2}\right\Vert_p^q+\left\Vert\dfrac{f-g}{2}\right\Vert_p^q\leq\left(\dfrac{\Vert f\Vert_p^p+\Vert g\Vert_p^p}{2}\right)^{q-1}&&(p\leq 2).\\
\end{align*}
$$

Czhang271828

实际上, $l^p$ 的题设可以去掉.

命题. 若 $\mathbb R$上赋范空间 $(X,\Vert\cdot\Vert)$ 满足平行四边形恒等式
$$
\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2+\Vert y\Vert ^2)\quad (\forall x,y\in X),
$$
则 $(x,y):=\dfrac{1}{4}(\Vert x+y\Vert ^2-\Vert x- y\Vert ^2)$ 是良定义的内积.

注. 考虑平行四边形法则, 上述内积与 $(x,y):=\dfrac{1}{2}(\Vert x+y\Vert ^2-\Vert x\Vert ^2-\Vert y\Vert ^2)$ 等价.

证明. 可以验证 $(x,y)=(y,x)$, $\Vert x\Vert^2=(x,x)$, $(x,y)$ 对双边连续. 以下证明 $(x,y)$ 是良定义的 $\mathbb R$-双线性型. 考虑到连续性, 只需证明 $(x,y)$ 是良定义的 $\mathbb Q$-双线性型.

对任意 $x,y,z\in X$​, 下证明 $(x+y,z)=(x,z)+(y,z)$​ 即可. 注意到
$$
\begin{align*}
\Vert x+y+z\Vert^2&=2\Vert x+z\Vert^2+2\Vert y\Vert^2-\Vert x-y+z\Vert^2,\\
\Vert x+y+z\Vert^2&=2\Vert y+z\Vert^2+2\Vert x\Vert^2-\Vert y-x+z\Vert^2.
\end{align*}
$$
将上两式相加得
$$
\Vert x+y+z\Vert^2=\Vert x+z\Vert^2+\Vert y+z\Vert^2+\Vert x\Vert^2+\Vert y\Vert^2-\dfrac{\Vert x-y+z\Vert^2+\Vert y-x+z\Vert^2}{2}.
$$
同理,
$$
\Vert x+y-z\Vert^2=\Vert x-z\Vert^2+\Vert y-z\Vert^2+\Vert x\Vert^2+\Vert y\Vert^2-\dfrac{\Vert x-y-z\Vert^2+\Vert y-x-z\Vert^2}{2}.
$$
相减得
$$
\Vert x+y+z\Vert^2-\Vert x+y-z\Vert^2=\Vert y+z\Vert^2+\Vert x+z\Vert^2-\Vert y-z\Vert^2-\Vert x-z\Vert^2,
$$
进而有
$$
\begin{align*}
(x+y,z)&=\dfrac{1}{4}(\Vert x+y+z\Vert^2-\Vert x+y-z\Vert^2)\\
&=\dfrac{1}{4}(\Vert y+z\Vert^2+\Vert x+z\Vert^2-\Vert y-z\Vert^2-\Vert x-z\Vert^2)\\
&=(x,z)+(y,z).
\end{align*}
$$
不难推得 $(x,y)$ 是 $\mathbb Q$-双线性型, 从而是内积.

Czhang271828

Czhang271828 发表于 2023-1-6 15:41
实际上, $l^p$ 的题设可以去掉.

命题. 若 $\mathbb R$上赋范空间 $(X,\Vert\cdot\Vert)$ 满足平行四边形 ...

命题. 若 $\mathbb C$上赋范空间 $(X,\Vert\cdot\Vert)$ 满足平行四边形恒等式
$$
\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2+\Vert y\Vert ^2)\quad (\forall x,y\in X),
$$
则 $\displaystyle (x,y):=\dfrac{1}{4}\sum_{s=0}^3i^s\Vert x+i^sy\Vert ^2$ 是良定义的内积.

证明. 可以验证 $(x,y)=\overline{(y,x)}$, $\Vert x\Vert^2=(x,x)$, $(x,y)$ 对双边连续. 以下只需证明 $(x,y)$ 是良定义的 $(x,y)$ 是良定义的 $\mathbb Q(i)$-一个半线性型.

搬运上楼中对平行四边形等式的变形, 得
$$
\begin{align*}
&\Vert x+y+z\Vert^2-\Vert x+y-z\Vert^2=\Vert y+z\Vert^2+\Vert x+z\Vert^2-\Vert y-z\Vert^2-\Vert x-z\Vert^2,\\
&\Vert x+y+iz\Vert^2-\Vert x+y-iz\Vert^2=\Vert y+iz\Vert^2+\Vert x+iz\Vert^2-\Vert y-iz\Vert^2-\Vert x-iz\Vert^2.
\end{align*}
$$
后式乘以 $i$​ 再加前式, 得
$$
\sum_{s=0}^3i^s\Vert x+y+i^sz\Vert^2=\sum_{s=0}^3i^s\Vert x+i^sz\Vert^2+\sum_{s=0}^3i^s\Vert y+i^sz\Vert^2,
$$
从而 $(x+y,z)=(x,z)+(y,z)$, 即 $(x,y)$ 是 $\mathbb Q$​-双线性的. 注意到
$$
\begin{align*}
(ix,y)&=\sum_{s=0}^3i^s\Vert ix+i^sy\Vert^2=\sum_{s=0}^3i^{s+1}\Vert x+i^sy\Vert^2=i(x,y),\\
(x,iy)&=\sum_{s=0}^3i^s\Vert x+i^{s+1}y\Vert^2=\sum_{s=0}^3i^{s-1}\Vert x+i^sy\Vert^2=-i(x,y).
\end{align*}
$$
从而 $(ix,y)=i(x,y)=(x,-iy)$, 结合 $\mathbb Q$-双线性性可知 $(x,y)$ 是良定义的 $\mathbb Q(i)$-一个半线性型.

hbghlyj

加了标签才发现3#与2个月之前的较粗糙的这个帖子是相同的.

hbghlyj

Czhang271828 发表于 2023-1-6 08:41
实际上, $l^p$ 的题设可以去掉.

命题. 若 $\mathbb R$上赋范空间 $(X,\Vert\cdot\Vert)$ 满足平行四边形 ...

特征为2的域上,不能除以4,就没有这个公式,那么二次型还能确定双线性型吗