How to rewrite Tanaka's formula in terms of right derivative and left local time? In Revuz and Yor's book, we have from Tanaka that $$|X_t -a| = |X_0 -a|+ \int_0^t{\text{sgn}(X_s-a)dX_s}+L_t^a,$$ $$(X_t -a)^+ = (X_0 -a)^++ \int_0^t{\mathbb{1}_{X_s>a}dX_s}+\frac{1}{2}L_t^a,$$ $$(X_t -a)^- = (X_0 -a)^-+ \int_0^t{-\mathbb{1}_{X_s \leq a}dX_s}+\frac{1}{2}L_t^a.$$ Here we have the integrand being the left-hand derivative of the initial function and $L_t^a$ being the right-hand local time. It states it is possible to rewrite these in terms of the integrand being the right-hand derivative and thus $L_t^a$ being the left-hand local time, how would you go about doing this?
Essentially what I'm asking is, the left-hand derivative of $f(x)=(x -a)^+$ is equal to $\mathbb{1}_{x>a}$, what is the right-hand derivative? And the left-hand derivative of $f(x)=(x -a)^-$ is equal to $-\mathbb{1}_{x \leq a}$, what is the right-hand derivative?
 A: The left-hand derivative of $f(x)$ is
$$
f'_-(x)=\lim_{y\uparrow x}\frac{f(x)-f(y)}{x-y}\,.
$$
The right-hand derivative of $f(x)$ is
$$
f'_+(x)=\lim_{y\downarrow x}\frac{f(y)-f(x)}{y-x}\,.
$$
For $f(x)=(x-a)^\pm$ the only slightly tricky point is $x=a\,.$
For $f(x)=(x-a)^+$we have obviously $f(a)=0$ and $f(y)=0$ for all $y<a\,.$ Therefore , $f'_-(a)=0\,.$ Applying similar considerations:
$$
f'_-(x)=1_{\{x>a\}}\,,\quad f'_+(x)=1_{\{x\ge a\}}\,.
$$
Likewise, for $f(x)=(x-a)^-$ we have
$$
f'_-(x)=-1_{\{x\le a\}}\,,\quad f'_+(x)=-1_{\{x<a\}}\,.
$$
The Tanaka formulas you quoted from Revuz & Yor use the following definitions:
\begin{align}
{\rm sgn}(x)&=\left\{\begin{array}{cc}+1,&x>1;\\-1,&x\le 1;\end{array}\right.\\
L_t^a(X)&=\lim_{\varepsilon\downarrow 0}\frac{1}{\varepsilon}\int_0^t1_{[a,a+\varepsilon[}(X_s)\,d\langle X,X\rangle_s\,.\text{(see Corr. VI (1.9))}
\end{align}
They don't seem to give this a name but you call that right-hand local time. Fine.
In Exercise VI (1.25) they introduce
\begin{align}
\widetilde{\rm sgn}(x)&=\left\{\begin{array}{cc}+1,&x>1;\\0,&x=0;\\-1,&x\le 1;\end{array}\right.\\
\tilde{L}_t^a(X)&=\lim_{\varepsilon\downarrow 0}\frac{1}{\varepsilon}\int_0^t1_{]a-\varepsilon,a+\varepsilon[}(X_s)\,d\langle X,X\rangle_s\,.
\end{align}
which they call symmetric local time. In terms of these the Tanaka formula is (for convex $f$)
\begin{align}\tag{2}
f(X_t)=f(X_0)+\int_0^t\frac{1}{2}(f_+'+f_-')\,dX_s+\frac{1}{2}\int\tilde{L}_tf''(da)\,.
\end{align}
Hint: To get the Tanaka formula in which only $f'_+$ appears just subtract
from (2) half of the Tanaka formula in which only $f_-'$ appears (the one you are familiar with). This should lead to an expression of the left-hand local time also.
