I need a hint (not a complete solution) of the following problem:
EDIT: when I was posting the question, I found I suddenly got it, so I need some verification.
Suppose $F$ is a complex-valued function of bounded variation on $[a,b]$. Then the total variation $T_F(a,b)\ge\int_a^b\lvert F'(x)\rvert\,dx$.
If $F$ is real-valued, then the problem is somewhat easier. Suppose the positive and the negative variation of $F$ on $[a,x]$ is $G(x),H(x)$, then $T_F(a,x)=G(x)+H(x)$ and $F(x)=G(x)-H(x)$, hence
$$\int_a^b\lvert F'(x)\rvert\,dx\le\int_a^bG'(x)\,dx+\int_a^bH'(x)\,dx\le (G(b)-G(a))+(H(b)-H(a))=T_F(a,b)$$
We have used the following lemma:
Suppose $F$ is increasing on $[a,b]$, then $\int_a^bF'(x)\,dx\le F(b)-F(a)$.
It could be proved as follows:
First, extend $F$ to $[a,+\infty)$, such that $f(x)=f(b),\;\forall x\ge b$. By Fatou's lemma,
\begin{align}
\int_a^bF'(x)\,dx
&=\int_a^b\lim_{n\to\infty}\frac{f(x+1/n)-f(x)}n\,dx\\
&\le\liminf_{n\to\infty}n\left(\int_{a+1/n}^{b+1/n}f-\int_a^bf\right)\\
&=\lim_{n\to\infty}n\left(\int_b^{b+1/n}f-\int_a^{a+1/n}f\right)\\
&=F(b)-F(a^+)
\end{align}
I cannot apply the same technique to telescope the terms when $F$ is complex-valued.
Note that when $F$ is absolutely continuous, then $T_F(a,b)\ge\int_a^b\lvert F'\rvert$ (in fact, $T_F(a,b)=\int_a^b$). The following argument should work:
Denote $\lVert f\rVert=\lVert f\rVert_{L^1}=\int_a^b\lvert F'\rvert$. Since $F$ is AC, therefore of BC, hence $F'\in L^1$, fix $\epsilon>0$, we could choose step function $g$ s.t. $\lVert F'-g\rVert\le\epsilon$, and supose $g$ induces the partition $a=t_0<\dotsb<t_N=b$, we have
$$\lVert F'\rVert\le\lVert F'-g\rVert+\lVert g\rVert\le\epsilon+\sum_{k=1}^N\left\lvert\int_{t_{k-1}}^{t_k}g\right\rvert\le2\epsilon+\sum_{k=1}^N\left\lvert\int_{t_{k-1}}^{t_k}F'\right\rvert\le2\epsilon+T_F(a,b)$$
It needs some explanation on the last step, i.e.:
If $F$ is of BV, then $\lvert\int_a^bF'\rvert\le\lvert F(b)-F(a)\rvert$. By the real case, the real part and the imaginery part satisfy the preceding inequality, and note that $\lvert a\rvert\le\lvert a'\rvert, \lvert b\rvert\le\lvert a'\rvert\implies \lvert a+bi\rvert\le\lvert a'+b'i\rvert$.
Best Answer
Let $TF(x) = T_F(a,x)$. For a point $x_0\in [a,b]$ where both $F$ and $TF$ are differentiable, we have $TF'(x_0)\ge |F'(x_0)|$. Hence $$T_F(a,b)\ge \int^{b}_{a} (TF)'{\rm d}x\ge \int^{b}_{a} |F'|{\rm d}x.$$ Note that this works when the range is any finite-dimensional real normed space, in which case every BV function is almost everywhere differentiable.