Bounded variation gives one-sided Dirichlet integrability

By [L1], write $u=P_u-N_u$ with $P_u(0)=N_u(0)=0$ and $P_u,N_u$ nondecreasing. Let $p:={\inf }_{0<t\le \delta }{P}_u(t)={\lim }_{t\downarrow 0}{P}_u(t),q:={\inf }_{0<t\le \delta }{N}_u(t)={\lim }_{t\downarrow 0}{N}_u(t)\mathrm{.}$ Since $u(t)\to 0$, one has $p-q=0$. If $p=q>0$, define $\tilde{P}(0)=\tilde{N}(0)=0$ and $\tilde{P}(t)=P_u(t)-p$, $\tilde{N}(t)=N_u(t)-p$ for $t>0$. Then $\tilde{P},\tilde{N}$ are again nondecreasing, nonnegative, normalized, and satisfy $u=\tilde{P}-\tilde{N}$, but $\tilde{P}(t)<P_u(t)$ for every $t>0$, contradicting the minimality in [L1]. Hence $p=q=0$, so $P_u(t)\to 0,N_u(t)\to 0,{\mathrm{Var}}_{[0,t]}(u)=P_u(t)+N_u(t)\to 0$ as $t\downarrow 0$ by [L2].

For $0\le t\le \delta$, define $H_N(t):={\int }_0^t{D}_N(s)ds\mathrm{.}$ Using [L3] and the change of variables $y=(2N+1)\pi s$, $H_N(t)={\int }_0^{(2N+1)\pi t}\sin y\cdot g_N(y)dy,$ where $g_N(y):=\frac{1}{(2N+1)\pi \sin (y\mathrm{/}(2N+1))}\mathrm{.}$ If $(2N+1)\pi t\le 1$, then $0<y\mathrm{/}(2N+1)\le 1<2$, so [L5] and $\mathrm{\mid }\sin y\mathrm{\mid }\le y$ give $\mathrm{\mid }\sin y{g}_N(y)\mathrm{\mid }\le 3\mathrm{/}\pi$ on $[0,(2N+1)\pi t]$, hence $\mathrm{\mid }{H}_N(t)\mathrm{\mid }\le 3\mathrm{/}\pi$. If $(2N+1)\pi t>1$, the same estimate controls the part on $[0,1]$, while on $[1,(2N+1)\pi t]$ the function $g_N$ is positive and decreasing because $0<y\mathrm{/}(2N+1)\le \pi \mathrm{/}2$. Bonnet's theorem [L4] therefore gives a point $\xi$ with ${\int }_1^{(2N+1)\pi t}\sin y{g}_N(y)dy=g_N(1){\int }_1^{\xi }\sin ydy+g_N((2N+1)\pi t){\int }_{\xi }^{(2N+1)\pi t}\sin ydy\mathrm{.}$ The absolute value is at most $2g_N(1)+2g_N((2N+1)\pi t)\le 4g_N(1)\le \frac{12}{\pi }$ by monotonicity of $g_N$ and [L5]. Thus $\mathrm{\mid }{H}_N(t)\mathrm{\mid }\le \frac{15}{\pi }(0\le t\le \delta ,N\ge 0)\mathrm{.}$

Fix $\epsilon >0$. By step 1.1, choose $\eta \in (0,\delta )$ such that ${\mathrm{Var}}_{[0,\eta ]}(u)<\frac{\pi \epsilon }{60}\mathrm{.}$ Applying [L4] to the monotone functions $P_u$ and $N_u$ on $[0,\eta ]$, and using $P_u(0)=N_u(0)=0$, step 1.2 yields $\mid {\int }_0^{\eta }{P}_u(t)D_N(t)dt\mid \le \frac{30}{\pi }{P}_u(\eta ),\mid {\int }_0^{\eta }{N}_u(t)D_N(t)dt\mid \le \frac{30}{\pi }{N}_u(\eta )\mathrm{.}$ Therefore $\mid {\int }_0^{\eta }u(t)D_N(t)dt\mid \le \frac{30}{\pi }(P_u(\eta )+N_u(\eta ))=\frac{30}{\pi }{\mathrm{Var}}_{[0,\eta ]}(u)<\frac{\epsilon }{2}\mathrm{.}$

On $[\eta ,\delta ]$, define $\psi (t):={\mathbf{1}}_{[\eta ,\delta ]}(t)\frac{u(t)e^{i\pi t}}{\sin (\pi t)}\mathrm{.}$ Because $\sin (\pi t)$ is bounded away from $0$ on $[\eta ,\delta ]$ and $u$ is bounded on the compact interval $[\eta ,\delta ]$, one has $\psi \in L^1(\mathbb{T})$. By [L3], ${\int }_{\eta }^{\delta }u(t)D_N(t)dt=\mathrm{Im}{\int }_0^1\psi (t)e^{2\pi iNt}dt=\mathrm{Im}\widehat{\psi }(-N)\mathrm{.}$

By [L6], $\widehat{\psi }(-N)\to 0$. So step 3.1 gives ${\int }_{\eta }^{\delta }u(t)D_N(t)dt\to 0.$ Choose $N_0$ such that the absolute value of this integral is below $\epsilon \mathrm{/}2$ for every $N\ge N_0$. Combining with step 2.1 shows that, for $N\ge N_0$, $\mid {\int }_0^{\delta }u(t)D_N(t)dt\mid <\epsilon \mathrm{.}$ Since $D_N(t)=\sin ((2N+1)\pi t)\mathrm{/}\sin (\pi t)$ on $(0,\delta ]$ by [L3], the stated limit follows.