The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space

Let $T>0$ and $d$ be as in the definition of the Lebesgue space $L^{2}([0,T];\mathbb{R}^{d})$, and adopt all of the notation of that definition, including $\mathcal{B}_{[0,T]}$, $\lambda_{[0,T]}$, $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$, the relation $\sim$, the pairing $\langle\cdot,\cdot\rangle_{L^{2}}$, the norm $\lVert\cdot\rVert_{L^{2}}$, and the distance $d_{L^{2}}$. Write $P_{T}$ for the set function $P_{T}(E)=T^{-1}\lambda_{[0,T]}(E)$ on $\mathcal{B}_{[0,T]}$, which is a probability measure by claim 1 of the restricted Lebesgue toolkit. Then the following hold.

1. (The integrands are measurable, and the pairing converges absolutely.) For all $u,v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ the functions $t\mapsto|u(t)|^{2}$ and $t\mapsto u(t)\cdot v(t)$ are $\mathcal{B}_{[0,T]}$-measurable, the second is integrable, and

$$\int_{[0,T]}|u\cdot v|\,d\lambda_{[0,T]}\le\tfrac{1}{2}\bigl(\lVert[u]\rVert_{L^{2}}^{2}+\lVert[v]\rVert_{L^{2}}^{2}\bigr) .$$

2. ($\mathcal{L}^{2}$ is a real vector space.) If $u,v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ and $c$ is real, then $u+v$ and $cu$ belong to $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$; with the pointwise operations this set is a real vector space whose zero element is the constant map $0$.

3. (The quotient is well formed.) The relation $\sim$ is an equivalence relation on $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$; for $u,v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ one has $u\sim v$ if and only if $\int_{[0,T]}|u-v|^{2}\,d\lambda_{[0,T]}=0$; and if $u\sim u'$ and $v\sim v'$ then $u+v\sim u'+v'$ and $cu\sim cu'$ for every real $c$. Consequently the operations $[u]+[v]=[u+v]$ and $c[u]=[cu]$ are well defined and make $L^{2}([0,T];\mathbb{R}^{d})$ a real vector space with zero element $[0]$.

4. (The pairing is a well-defined inner product.) If $u\sim u'$ and $v\sim v'$ then $\int_{[0,T]}u\cdot v\,d\lambda_{[0,T]}=\int_{[0,T]}u'\cdot v'\,d\lambda_{[0,T]}$, so $\langle\cdot,\cdot\rangle_{L^{2}}$ is well defined on $L^{2}([0,T];\mathbb{R}^{d})$. It is symmetric and linear in each argument, and satisfies $\langle[u],[u]\rangle_{L^{2}}=\lVert[u]\rVert_{L^{2}}^{2}\ge0$, with $\lVert[u]\rVert_{L^{2}}=0$ if and only if $[u]=[0]$.

5. (Cauchy-Schwarz and triangle inequalities.) For all $u,v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$,

$$\bigl|\langle[u],[v]\rangle_{L^{2}}\bigr|\le\lVert[u]\rVert_{L^{2}}\,\lVert[v]\rVert_{L^{2}},\qquad \lVert[u]+[v]\rVert_{L^{2}}\le\lVert[u]\rVert_{L^{2}}+\lVert[v]\rVert_{L^{2}},$$

and $\lVert c[u]\rVert_{L^{2}}=|c|\,\lVert[u]\rVert_{L^{2}}$ for every real $c$.

6. (The distance is a metric.) $\bigl(L^{2}([0,T];\mathbb{R}^{d}),d_{L^{2}}\bigr)$ is a metric space.

7. (Comparison with the mean-square norm.) For every $\mathcal{B}_{[0,T]}$-measurable $f:[0,T]\to[0,\infty]$ one has $\int_{[0,T]}f\,dP_{T}=T^{-1}\int_{[0,T]}f\,d\lambda_{[0,T]}$. A map $u:[0,T]\to\mathbb{R}^{d}$ belongs to $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ if and only if each component $u^{i}$ is a square-integrable random variable on the probability space $([0,T],\mathcal{B}_{[0,T]},P_{T})$, and in that case, with $\lVert\cdot\rVert_{2}$ the mean-square norm of that definition,

$$\lVert[u]\rVert_{L^{2}}^{2}=T\sum_{i=1}^{d}\lVert u^{i}\rVert_{2}^{2}.$$

Moreover $u\sim v$ holds if and only if $u^{i}$ and $v^{i}$ are equal $P_{T}$-almost surely for every $i\in\{1,\dots,d\}$.