[proofplan] All five parts are derived from the polarization identity $\langle M, N \rangle := \frac{1}{4}(\langle M + N \rangle - \langle M - N \rangle)$ and the corresponding properties of the diagonal quadratic variation $\langle M \rangle$. Part (i) follows from a uniqueness argument: if two processes satisfy the characterization, their difference is a continuous local martingale of finite variation starting at zero, hence identically zero. Parts (ii) and (iv) reduce to the corresponding properties of $\langle \cdot \rangle$ via polarization. Part (iii) follows from the Riemann-sum convergence of the diagonal quadratic variation combined with the bilinearity of the Riemann sums. Part (v) follows from the [Quadratic Variation Norm Formula](/theorems/2085) applied to $M + N$ and $M - N$. [/proofplan]

[step:Establish the polarization identity as the definition of covariation] Define the covariation of continuous local martingales $M$ and $N$ by \begin{align*} \langle M, N \rangle_t := \frac{1}{4}\bigl(\langle M + N \rangle_t - \langle M - N \rangle_t\bigr). \end{align*} This is well-defined because $M + N$ and $M - N$ are continuous local martingales (the space of continuous local martingales is a vector space), so their quadratic variations exist. [/step]

[step:Prove part (i): characterization of $\langle M, N \rangle$ via the local martingale property]We first show that $M_t N_t - \langle M, N \rangle_t$ is a continuous local martingale. Expanding via the polarization identity, \begin{align*} M_t N_t &= \frac{1}{4}\bigl((M_t + N_t)^2 - (M_t - N_t)^2\bigr). \end{align*} Therefore \begin{align*} M_t N_t - \langle M, N \rangle_t &= \frac{1}{4}\bigl((M_t + N_t)^2 - \langle M + N \rangle_t\bigr) - \frac{1}{4}\bigl((M_t - N_t)^2 - \langle M - N \rangle_t\bigr). \end{align*} By definition of quadratic variation, $(M + N)^2_t - \langle M + N \rangle_t$ and $(M - N)^2_t - \langle M - N \rangle_t$ are both continuous local martingales. A linear combination of continuous local martingales is a continuous local martingale, so $M_t N_t - \langle M, N \rangle_t$ is a continuous local martingale. The process $\langle M, N \rangle$ is continuous (as a linear combination of continuous processes) and of finite variation (as a difference of two increasing processes, each divided by $4$), and satisfies $\langle M, N \rangle_0 = \frac{1}{4}(\langle M + N \rangle_0 - \langle M - N \rangle_0) = 0$. **Uniqueness.** Suppose $A$ is another continuous finite variation process starting at zero such that $M_t N_t - A_t$ is a continuous local martingale. Then \begin{align*} A_t - \langle M, N \rangle_t = (M_t N_t - \langle M, N \rangle_t) - (M_t N_t - A_t) \end{align*} is a continuous local martingale. Since $A - \langle M, N \rangle$ is also of finite variation (as the difference of two finite variation processes) and starts at zero, the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem gives $A_t = \langle M, N \rangle_t$ for all $t \geq 0$ a.s.[/step]

[guided]The characterization says that $\langle M, N \rangle$ is the *unique* continuous finite variation process starting at zero that makes $MN - \langle M, N \rangle$ a continuous local martingale. This mirrors the characterization of the diagonal quadratic variation $\langle M \rangle$. The existence part uses the algebraic identity $ab = \frac{1}{4}((a+b)^2 - (a-b)^2)$ applied to $a = M_t$, $b = N_t$. This gives \begin{align*} M_t N_t = \frac{1}{4}(M_t + N_t)^2 - \frac{1}{4}(M_t - N_t)^2. \end{align*} Subtracting the polarization definition of $\langle M, N \rangle$ and rearranging, we express $M_t N_t - \langle M, N \rangle_t$ as a linear combination of the two processes $(M+N)^2 - \langle M+N \rangle$ and $(M-N)^2 - \langle M-N \rangle$, each of which is a continuous local martingale by the characterizing property of quadratic variation. For uniqueness, if $A$ is any other process with the same properties, then $A - \langle M, N \rangle$ is the difference of two continuous local martingales (namely $MN - \langle M, N \rangle$ and $MN - A$), hence a continuous local martingale. It is also of finite variation (as the difference of two finite variation processes) and starts at zero. The [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem then forces $A = \langle M, N \rangle$.[/guided]

[step:Prove part (ii): bilinearity and symmetry] **Symmetry.** From the polarization identity, \begin{align*} \langle N, M \rangle_t = \frac{1}{4}\bigl(\langle N + M \rangle_t - \langle N - M \rangle_t\bigr) = \frac{1}{4}\bigl(\langle M + N \rangle_t - \langle M - N \rangle_t\bigr) = \langle M, N \rangle_t, \end{align*} where we used $\langle N - M \rangle = \langle -(M - N) \rangle = \langle M - N \rangle$ (since $\langle \alpha X \rangle = \alpha^2 \langle X \rangle$ and $(-1)^2 = 1$). **Linearity in the first argument.** Let $\alpha, \beta \in \mathbb{R}$ and let $M, M', N$ be continuous local martingales. We show $\langle \alpha M + \beta M', N \rangle = \alpha \langle M, N \rangle + \beta \langle M', N \rangle$. Define $A_t := \alpha \langle M, N \rangle_t + \beta \langle M', N \rangle_t$. This is a continuous finite variation process starting at zero. We verify that $(\alpha M + \beta M')_t N_t - A_t$ is a continuous local martingale: \begin{align*} (\alpha M_t + \beta M'_t) N_t - A_t &= \alpha(M_t N_t - \langle M, N \rangle_t) + \beta(M'_t N_t - \langle M', N \rangle_t). \end{align*} Each term in parentheses is a continuous local martingale (by the characterization in part (i)), and a linear combination of continuous local martingales is a continuous local martingale. By uniqueness (part (i)), $A = \langle \alpha M + \beta M', N \rangle$. Linearity in the second argument follows from symmetry and linearity in the first argument. [/step]

[step:Prove part (iii): Riemann-sum convergence]Define the Riemann-sum approximations \begin{align*} \langle M, N \rangle^{(n)}_t := \sum_{i=1}^{\lfloor 2^n t \rfloor} (M_{i \cdot 2^{-n}} - M_{(i-1) \cdot 2^{-n}})(N_{i \cdot 2^{-n}} - N_{(i-1) \cdot 2^{-n}}). \end{align*} Using the identity $ab = \frac{1}{4}((a+b)^2 - (a-b)^2)$ with $a = M_{i \cdot 2^{-n}} - M_{(i-1) \cdot 2^{-n}}$ and $b = N_{i \cdot 2^{-n}} - N_{(i-1) \cdot 2^{-n}}$, we obtain \begin{align*} \langle M, N \rangle^{(n)}_t = \frac{1}{4}\bigl(\langle M + N \rangle^{(n)}_t - \langle M - N \rangle^{(n)}_t\bigr), \end{align*} where $\langle X \rangle^{(n)}_t := \sum_{i=1}^{\lfloor 2^n t \rfloor} (X_{i \cdot 2^{-n}} - X_{(i-1) \cdot 2^{-n}})^2$ denotes the diagonal Riemann-sum approximation. By the u.c.p. convergence of the diagonal quadratic variation Riemann sums (applied to the continuous local martingales $M + N$ and $M - N$), \begin{align*} \langle M + N \rangle^{(n)}_t &\xrightarrow{\text{u.c.p.}} \langle M + N \rangle_t, \\ \langle M - N \rangle^{(n)}_t &\xrightarrow{\text{u.c.p.}} \langle M - N \rangle_t. \end{align*} Since u.c.p. convergence is preserved under linear combinations (if $X^{(n)} \xrightarrow{\text{u.c.p.}} X$ and $Y^{(n)} \xrightarrow{\text{u.c.p.}} Y$, then $\alpha X^{(n)} + \beta Y^{(n)} \xrightarrow{\text{u.c.p.}} \alpha X + \beta Y$ for any $\alpha, \beta \in \mathbb{R}$), \begin{align*} \langle M, N \rangle^{(n)}_t = \frac{1}{4}\bigl(\langle M + N \rangle^{(n)}_t - \langle M - N \rangle^{(n)}_t\bigr) \xrightarrow{\text{u.c.p.}} \frac{1}{4}\bigl(\langle M + N \rangle_t - \langle M - N \rangle_t\bigr) = \langle M, N \rangle_t. \end{align*}[/step]

[guided]The Riemann-sum convergence for the covariation is an immediate consequence of the corresponding result for the diagonal quadratic variation, via the polarization trick. The key algebraic observation is that the product of increments factors through sums and differences of squares: \begin{align*} \Delta M_i \cdot \Delta N_i = \frac{1}{4}\bigl((\Delta M_i + \Delta N_i)^2 - (\Delta M_i - \Delta N_i)^2\bigr) = \frac{1}{4}\bigl((\Delta(M+N)_i)^2 - (\Delta(M-N)_i)^2\bigr). \end{align*} Summing over $i$ gives $\langle M, N \rangle^{(n)}_t = \frac{1}{4}(\langle M+N \rangle^{(n)}_t - \langle M-N \rangle^{(n)}_t)$. Each diagonal approximation converges u.c.p. to the corresponding quadratic variation. The u.c.p. mode of convergence ($\sup_{s \leq t} |X^{(n)}_s - X_s| \xrightarrow{\mathbb{P}} 0$ for each $t$) is preserved under fixed linear combinations because $\sup_{s \leq t} |\alpha X^{(n)}_s + \beta Y^{(n)}_s - \alpha X_s - \beta Y_s| \leq |\alpha| \sup_{s \leq t} |X^{(n)}_s - X_s| + |\beta| \sup_{s \leq t} |Y^{(n)}_s - Y_s|$, and both terms on the right converge to zero in probability.[/guided]

[step:Prove part (iv): stopped processes satisfy $\langle M^T, N^T \rangle_t = \langle M^T, N \rangle_t = \langle M, N \rangle_{t \wedge T}$] **First identity: $\langle M, N \rangle_{t \wedge T} = \langle M^T, N^T \rangle_t$.** By the polarization identity and the [Quadratic Variation of a Stopped Process](/theorems/2083) theorem, \begin{align*} \langle M^T, N^T \rangle_t &= \frac{1}{4}\bigl(\langle M^T + N^T \rangle_t - \langle M^T - N^T \rangle_t\bigr) \\ &= \frac{1}{4}\bigl(\langle (M + N)^T \rangle_t - \langle (M - N)^T \rangle_t\bigr) \\ &= \frac{1}{4}\bigl(\langle M + N \rangle_{t \wedge T} - \langle M - N \rangle_{t \wedge T}\bigr) \\ &= \langle M, N \rangle_{t \wedge T}, \end{align*} where the second equality uses $(M + N)^T = M^T + N^T$ and $(M - N)^T = M^T - N^T$ (stopping commutes with linear combinations), and the third uses the [Quadratic Variation of a Stopped Process](/theorems/2083) theorem applied to $M + N$ and $M - N$. **Second identity: $\langle M^T, N \rangle_t = \langle M, N \rangle_{t \wedge T}$.** Consider the process $M^T_t N_t - \langle M, N \rangle_{t \wedge T}$. We show this is a continuous local martingale. Write \begin{align*} M^T_t N_t - \langle M, N \rangle_{t \wedge T} = M^T_t N_t - \langle M^T, N^T \rangle_t, \end{align*} using the first identity. But this is *not* quite what we need, since the right-hand side involves $N^T$ rather than $N$. Instead, we use the characterization approach. The process $B_t := \langle M, N \rangle_{t \wedge T}$ is a continuous finite variation process starting at zero (by the same argument as in the stopped quadratic variation proof). We need to show $M^T_t N_t - B_t$ is a continuous local martingale. Since $M_t N_t - \langle M, N \rangle_t$ is a continuous local martingale, stopping at $T$ gives that $(MN)_{t \wedge T} - \langle M, N \rangle_{t \wedge T} = M^T_t N^T_t - B_t$ is a continuous local martingale. Now write \begin{align*} M^T_t N_t - B_t = (M^T_t N^T_t - B_t) + M^T_t(N_t - N^T_t). \end{align*} The first term is a continuous local martingale. For the second term, note that $N_t - N^T_t = N_t - N_{t \wedge T}$, which equals zero for $t \leq T$ and $N_t - N_T$ for $t > T$. The process $M^T_t$ is constant for $t \geq T$ (equal to $M_T$), so \begin{align*} M^T_t(N_t - N^T_t) = M_T(N_t - N_T) \mathbb{1}_{\{t > T\}}. \end{align*} This is a continuous local martingale (it is the stochastic integral of the previsible process $M_T \mathbb{1}_{\{s > T\}}$ with respect to $N$). Therefore $M^T_t N_t - B_t$ is a continuous local martingale. By uniqueness (part (i)), $B = \langle M^T, N \rangle$, giving $\langle M^T, N \rangle_t = \langle M, N \rangle_{t \wedge T}$. [/step]

[step:Prove part (v): the square-integrable case and the inner product formula]Suppose $M, N \in \mathcal{M}^2_c$. By the [Quadratic Variation Norm Formula](/theorems/2085) applied to $M + N$ and $M - N$ (both of which lie in $\mathcal{M}^2_c$ since $\mathcal{M}^2_c$ is a vector space), \begin{align*} (M + N)^2_t - \langle M + N \rangle_t \quad \text{and} \quad (M - N)^2_t - \langle M - N \rangle_t \end{align*} are both uniformly integrable martingales. Since $M_t N_t - \langle M, N \rangle_t = \frac{1}{4}\bigl((M+N)^2_t - \langle M+N \rangle_t\bigr) - \frac{1}{4}\bigl((M-N)^2_t - \langle M-N \rangle_t\bigr)$, and a linear combination of uniformly integrable martingales is a uniformly integrable martingale, $M_t N_t - \langle M, N \rangle_t$ is a uniformly integrable martingale. For the inner product formula, the [Quadratic Variation Norm Formula](/theorems/2085) gives \begin{align*} \|M + N - (M_0 + N_0)\|_{\mathcal{M}^2}^2 &= \mathbb{E}[\langle M + N \rangle_\infty], \\ \|M - N - (M_0 - N_0)\|_{\mathcal{M}^2}^2 &= \mathbb{E}[\langle M - N \rangle_\infty]. \end{align*} Subtracting and dividing by $4$, \begin{align*} \frac{1}{4}\bigl(\|M + N - (M_0 + N_0)\|_{\mathcal{M}^2}^2 - \|M - N - (M_0 - N_0)\|_{\mathcal{M}^2}^2\bigr) = \frac{1}{4}\bigl(\mathbb{E}[\langle M + N \rangle_\infty] - \mathbb{E}[\langle M - N \rangle_\infty]\bigr) = \mathbb{E}[\langle M, N \rangle_\infty]. \end{align*} The left-hand side is the polarization of the $\mathcal{M}^2$-norm applied to $M - M_0$ and $N - N_0$, which equals the inner product $\langle M - M_0, N - N_0 \rangle_{\mathcal{M}^2}$ (using the standard polarization identity for a Hilbert space inner product: $\langle u, v \rangle = \frac{1}{4}(\|u + v\|^2 - \|u - v\|^2)$). Therefore \begin{align*} \langle M - M_0, N - N_0 \rangle_{\mathcal{M}^2} = \mathbb{E}[\langle M, N \rangle_\infty]. \end{align*}[/step]

[guided]Part (v) connects the stochastic covariation $\langle M, N \rangle$ to the Hilbert space inner product on $\mathcal{M}^2_c$. The argument uses polarization twice: once at the stochastic level (to define $\langle M, N \rangle$ from $\langle M \rangle$) and once at the Hilbert space level (to recover the inner product from the norm). For the uniform integrability of $MN - \langle M, N \rangle$: since $M, N \in \mathcal{M}^2_c$, both $M + N$ and $M - N$ are in $\mathcal{M}^2_c$, and the [Quadratic Variation Norm Formula](/theorems/2085) guarantees that $(M \pm N)^2 - \langle M \pm N \rangle$ are uniformly integrable martingales. Taking the difference and dividing by $4$ gives the result. For the inner product formula, the Hilbert space polarization identity states that in any real inner product space, $\langle u, v \rangle = \frac{1}{4}(\|u+v\|^2 - \|u-v\|^2)$. Applying this to $u = M - M_0$ and $v = N - N_0$ in $\mathcal{M}^2_c$: \begin{align*} \langle M - M_0, N - N_0 \rangle_{\mathcal{M}^2} &= \frac{1}{4}\bigl(\|M - M_0 + N - N_0\|^2 - \|M - M_0 - N + N_0\|^2\bigr). \end{align*} The norm formula gives $\|M - M_0 + N - N_0\|^2 = \|(M+N) - (M_0 + N_0)\|^2 = \mathbb{E}[\langle M+N \rangle_\infty]$ and similarly for $M - N$. Substituting yields $\langle M - M_0, N - N_0 \rangle_{\mathcal{M}^2} = \frac{1}{4}(\mathbb{E}[\langle M+N \rangle_\infty] - \mathbb{E}[\langle M-N \rangle_\infty]) = \mathbb{E}[\langle M, N \rangle_\infty]$. This formula is the foundation for the Ito isometry: it says that the map $M \mapsto \langle M \rangle$ intertwines the Hilbert space structure of $\mathcal{M}^2_c$ with the $L^1$ structure of quadratic variations.[/guided]