Let $f : [a, b] \rightarrow R $ of bounded variation. Let $$p_{f}(x) =\frac{1}{2}[V_{f}(x) + f(x)-f(a)]$$ and $$q_{f }(x) =\frac{1}{2}[V_{f}(x) - f(x) + f(a)]$$ be the positive and negative variations of $f$, respectively. Suppose that $p_{1}, q_{1} : [a, b] \rightarrow \mathbb{R} $ are increasing functions such that $f = p_{1}-q_{1}$. Show that if $a\leq x < y \leq b$, then $$p_{f}(x)-p_{f}(y) \leq p_{1}(x)-p_{1}(y)$$ and $$q_{f}(x)-q_{f}(y)\leq q_{1}(x)-q_{1}(y)$$ Conclude that $V_{p_{f}}(a, b)\leq V_{p_{1}}(a, b)$ and $V_{q_{f}}(a, b) \leq V_{q_{1}}(a, b)$.
We did the following $$p_{f}(x)-p_{f}(y)=\frac{1}{2}[V_{f}(x)-V_{f}(y)+f(x)-f(y)]=\frac{1}{2}[-V_{f}(x,y)+f(x)-f(y)] $$ As $$f(y)-f(x)\leq V_{f}(x,y) \Leftrightarrow -V_{f}(x,y)\leq f(x)-f(y) $$ then $$\frac{1}{2}[-V_{f}(x,y)+f(x)-f(y)]\leq f(x)-f(y)=p_{1}(x)-p_{1}(y)+q_{1}(y)-q_{1}(x) $$ Note that $p_{1}(x)-p_{1}(y)\leq 0\leq q_{1}(y)-q_{1}(x)$
We haven't been able to conclude what we want to demonstrate. Could someone help us?
I think you invert $x$ and $y$ in your inequality.
I guess you may look at the problem this way : instead of your expression of $p_f$, use its definition $$p_f(x)= \sup_{(x_i) \text{ subdivision of } [a,x]} \max (0,f(x_{i+1})-f(x_i)).$$ Then for $x<y$, $$p_f(y)-p_f(x)=\sup_{(x_i) \text{ subdivision of } [x,y]} \max (0,f(x_{i+1})-f(x_i)),$$ but $\max (0,f(x_{i+1})-f(x_i)) \leqslant p_1(x_{i+1})-p_1(x_i)$, and you can conclude.
Do the same for the inequality with $q$.