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The Itô formula decomposes the smooth transform into a continuous local martingale and a finite-variation process:If $f(M)$ has finite variation, then $N$ also has finite variation. A continuous finite-variation local martingale is constant, so $N=0$ andIt remains to show that the second-order term vanishes; the preceding observation alone does not establish this.

Let $Z=\{a:f^{\prime }(a)=0\}$. By the occupation-times formula for the local time of a semimartingale,Therefore $L_t^a(M)=0$ for almost every $a\notin Z$. Also $f^{\prime \prime }(a)=0$ for almost every $a\in Z$: if $f^{\prime }(a)=0$ and $f^{\prime \prime }(a)\ne 0$, continuity of $f^{\prime \prime }$ makes $f^{\prime }$ strictly monotone in a neighborhood of $a$, so that zero is isolated. Such isolated zeros form a countable set. Localizing to bounded paths and finite quadratic variation, then using the occupation-times formula once more, givesBoth terms in the Itô formula now vanish. Apply this on all rational times and use continuity to obtain the simultaneous conclusionThis is the finite-variation smooth transform of a continuous local martingale property.

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

The initial-value convention is essential. A continuous finite-variation local martingale is constant, so the conclusion is $M_t=M_0$ up to indistinguishability of stochastic processes. The stated zero conclusion uses the usual zero-start convention $M_0=0$; without it, $M_t\equiv 1$ is a counterexample.

Put $N=M-M_0$. Let $V_t$ be its total-variation process, and stop when either $|N|$ or $V$ reaches $n$. The resulting process $N^{{\tau }_n}$ is a bounded continuous local martingale, hence a true martingale, and its total variation on every time interval is bounded by $n$. Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_k=t$ whose mesh tends to zero. Orthogonality of martingale increments givesPathwise,by uniform continuity of a continuous path on $[0,t]$. The sums are bounded by $2n^2$, so the dominated convergence theorem makes their expectations tend to zero. Hence $N_t^{{\tau }_n}=0$ almost surely. Take the countable intersection over rational $t$ and integer $n$, use path continuity, and then let $n\to \infty$. The stopping times tend to infinity because the original path and its finite variation are bounded on every compact time interval. Thus

The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; finite variation alone does not supply them. For example, $M=0$, $A_t=1_{\{t\ge 1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.

Define $f_n(x)={\int }_0^x\varphi (ny)dy$. Then $f_n\in C^2$, $|f_n^{\prime }|\le 1$ and $f_n^{\prime \prime }(x)=n{\varphi }^{\prime }(nx)\ge 0$. The Itô formula gives the canonical continuous semimartingale decompositionThe initial value is zero. The first term is a continuous martingale, since $M$ is bounded and the integrand is bounded. The second is continuous adapted finite variation, hence predictable. This decomposition is unique because a continuous finite-variation local martingale is constant. The phrase Doob-Meyer here must mean this semimartingale decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the Doob-Meyer decomposition theorem. Indeed $f_n^{\prime }(0)=-1$, so the bounded deterministic drift $A_t=t\wedge \delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.

The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,For a fixed horizon $T$, $\mathbb{E}[M{]}_T=\mathbb{E}{M}_T^2\le K^2$. The dominated convergence theorem for the finite measure $\mathbb{E}d[M]$ on $[0,T]$ givesThe Itô isometry, extended from simple predictable integrands by their density, and the Doob L2 maximal inequality now implyHence

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

For a continuous path $m$ of finite variation on $[0,t]$, every partition satisfiesUniform continuity makes the right side tend to zero as the mesh tends to zero. The quadratic variation of the given continuous local martingale $M$ is therefore identically zero.

Stop $M$ when its magnitude first reaches an integer $n$, so the stopped process is a bounded, square-integrable martingale. The square identity, or the Itô isometry, gives $\mathbb{E}{M}_{t\wedge {\tau }_n}^2=\mathbb{E}[M{]}_{t\wedge {\tau }_n}=0$. Hence it is zero at each rational time almost surely, and continuity extends this simultaneously to all times. Let $n$ tend to infinity. This provesThis is the continuous finite-variation local martingale is constant principle with initial value zero.

Use two standard results: for every continuous local martingale $U$, there is a unique continuous adapted increasing quadratic variation $[U]$, starting at zero, such that $U^2-U_0^2-[U]$ is a local martingale; and a continuous finite-variation local martingale is constant.

Define the quadratic covariation by the polarization identityThis is continuous, adapted, starts at zero, and is of finite variation on compact time intervals. Since $(M+N{)}^2-(M-N{)}^2=4MN$, subtraction of the two quadratic-variation identities shows that $MN-M_0{N}_0-[M,N]$ is a continuous local martingale. In the zero-start convention this is exactly $MN-[M,N]\in {\mathcal{M}}_{c,\mathrm{loc}}$.

If $A$ is another continuous adapted finite-variation process with $A_0=0$ and the same local martingale property, then $A-[M,N]$ is both a continuous local martingale and of finite variation. It is constant, and its initial value is zero, so it vanishes. Uniqueness is up to indistinguishability of stochastic processes.

The normalization $[M,N{]}_0=0$ is essential. If arbitrary nonzero initial constants are allowed in the martingale class and this normalization is omitted, adding a constant to the bracket preserves the claimed property. The usual definition of quadratic covariation always includes the zero initial value.

Use polarization identity of the quadratic variations:This is continuous and adapted, starts at zero, and has finite variation on every compact interval because it is a difference of two nondecreasing processes. The defining quadratic variation property gives that $(M+N{)}^2-[M+N]$ and $(M-N{)}^2-[M-N]$ are continuous local martingales; their difference divided by four is $MN-[M,N]$. This proves existence without assuming a product formula involving an as yet undefined quadratic covariation.

If $C$ and $D$ both satisfy the requirements, $C-D$ is a continuous finite-variation process and a local martingale. By continuous finite-variation local martingale is constant, it is constant; its initial value is zero, so $C=D$ up to indistinguishability of stochastic processes.

There is a minor initial-value convention in this characterization. The displayed uncentred local martingale identity is directly valid when $M_0=N_0=0$, or when the initial terms have the requisite integrability. With arbitrary finite initial values, centre the quadratic variation construction and use $MN-M_0{N}_0-[M,N]$, which starts at zero. Adding back $M_0{N}_0$ requires it to be integrable under the usual definition of a local martingale; the quadratic covariation itself is unaffected by centring.

The Itô formula applied to $F(t,x)=e^{t/2}\cos x$ gives exact cancellation of the drift:On every finite interval $[0,T]$, the Itô integral is square-integrable since $\mathbb{E}{\int }_0^T{e}^s{\sin }^2{B}_{s}ds\le e^T-1$. Therefore $e^{t/2}\cos B_t$ is a true martingale, and hence a local martingale.

For the other stochastic process, $d(B_t-t^2)=d{B}_t-2tdt$. If $B_t-t^2$ were a local martingale, its difference with $B$ would make the continuous finite-variation process $-t^2$ a local martingale, contradicting the fact that a continuous finite-variation local martingale is constant. Also $\mathbb{E}(B_t-t^2)=-t^2$ is not constant. Thus $B_t-t^2$ is neither a local martingale nor a martingale.

Define the deterministic function $f(t)=\mathbb{E}[M_t^2]$. The independent increments from part (i) show that $f$ is increasing and that $M_t^2-f(t)$ is a martingale. Mean-square continuity follows from path continuity and the Gaussian laws, so $f$ is continuous.

The Itô formula also says that $M_t^2-[M{]}_t$ is a local martingale. Their difference $[M{]}_t-f(t)$ is therefore a continuous finite-variation process that is also a local martingale. By the theorem that a continuous finite-variation local martingale is constant, and because the difference starts at zero,for all $t\ge 0$ almost surely.