---- $\mathbb{F}$ [[field|denotes]] $\mathbb{R}$ or $\mathbb{C}$. > [!theorem] Theorem. ([[nontrivial Lp spaces are Hilbert iff p=2]]) > Let $(X , \Sigma, \mu)$ be a nice enough [[measure|measure space]] and $1 \leq p \leq \infty$. The [[Lp-norm|space]] $L^{p}(\mu)$ is a [[Hilbert space]] if and only if $p=2$ or $\text{dim }L^{p}(\mu)=1$. In this case, the [[norm]] $\|\cdot\|_{2}$ is [[canonical norm induced by inner product|induced]] by the [[Lp duality|Hölder pairing]] $\langle f,g \rangle := \int f \overline{g} \, d\mu .$ ^theoremx > [!proof]+ Proof. ([[nontrivial Lp spaces are Hilbert iff p=2]]) > Recall that [[Lp spaces are Banach]]. So we don't need to worry about [[complete|completeness]], just the [[inner product]] structure. > > **Only if.** If $\|\cdot\|_{p}$ is [[canonical norm induced by inner product|induced]] by some [[inner product]] $\langle -,- \rangle_{p}$, then the the [[parallelogram law]] holds, stating $\|f+g\|^{2}_{p} + \|f-g\|^{2}_{p} =2 \|f\|^{2}_{p} + 2 \|g\|^{2}_{p}$ for all $f,g$ in the inner product space $(X, \langle -,- \rangle_{p})$. Take $E,F \in \Sigma$, chosen disjoint to entail $\mu(E \cup F)=\mu(E) + \mu(F)$. Noting that $\|\chi_{E}\|_{p}=\mu(E)^{1/p}$, $\|\chi_{F}\|_{p}=\mu (F)^{1/p}$ for all $1 \leq p \leq \infty$ ($1 / \infty=0$) we have > > $\begin{align} > \|\chi_{E} + \chi_{F}\|_{p}^{2} + \|\chi_{E}-\chi_{F}\|_{p}^{2} &= 2\big( \mu(E) + \mu(F) \big)^{2/p} \\ > 2 \|f\|_{p}^{2} + 2 \|g\|_{p}^{2} &= 2 \mu(E)^{2/p} + 2 \mu(F)^{2/p} > \end{align}$ > > and the parallelogram law thus stating $2\big( \mu(E) + \mu(F) \big)^{2/p} = 2 \mu(E)^{2/p} + 2 \mu(F)^{2/p}. \ (*)$ > Choose $E,F$ so that $\mu(E)=1=\mu(F)$. While $(*)$ does hold for $p=2$, it doesn't hold the other $p \in [1, \infty]$. For example, choose $E$, $F$ with unit [[measure]]; then $(*)$ gives the equation $2 \cdot 2^{2/p}=4$whose only solution in $[1, \infty]$ is $p=2$. > > **If.** We now show that taking $p=2$ indeed yields an [[inner product]]. [[dual exponent|Note that]] in this case $p'=2=p$. Using [[Hölder's inequality]], $\int f \overline{g} \, d\mu \leq \int |f \overline{g}| \, d\mu \leq \|f\|_{2} \|g\|_{2} <\infty$ > > for all $f,g \in L^{2}(\mu)$. Thus the formula $\langle f,g \rangle := \int f \overline{g} \, d\mu \ (f, g \in L^{2}(\mu))$ > indeed defines a map $L^{2}(\mu) \times L^{2}(\mu) \to \mathbb{F}$. > > Clearly $\sqrt{ \langle f, f \rangle }=\left( \int f ^{2} \, d\mu \right)^{1/2}=(\int |f|^{2} \, \, d\mu )^{1/2},$ > so we just have to show the proposed formula indeed defines an [[inner product space|inner product]] on $L^{2}(\mu)$. Positivity is clear and definiteness follows from [[integral|the property that]] the integral of a nonnegative function is zero if and only if the function is zero [[almost-everywhere]] (we're using $L^{2}$ not $\mathcal{L}^{2}$, so this is enough). Other conditions are immediate also from basic integral properties. ^1ab01d ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```