If $f\colon[a,b]\rightarrow \mathbb{R}$ is a continuous function, is it true that $\mathcal{W}(f,P) \leq V_f[a, b]$ for any partition $P$ of $[a,b]$?

Question

Let $f\colon[a,b]\rightarrow \mathbb{R}$ be a continuos function, is it true that $\mathcal{W}(f,P) \leq V_f[a, b]$ for any partition $P$ of $[a,b]$?

Where $V_f[a,b] = \sup \{ v_f(P): \text{P is a partition of} [a,b]\}$ is the total variation and $\mathcal{W}(f,P)$ is defined as follows: Let $f\colon[a,b]\rightarrow \mathbb{R}$ and $P= \{x_0,...,x_n\}$ a partition of $[a,b]$. For each $k \in \{1,...,n\}$, we define $M_k = \sup \{f(x):x\in[x_{k-1},x_k]\}$ and $m_k = \inf\{f(x):x\in[x_{k-1},x_k]\}$. Then $\mathcal{W}(f,P) = \sum_{k=1}^n M_k-m_k$.

Intuitively, I can't find any scenario where I would have $\mathcal{W}(f,P) > V_f[a, b]$, since I could just divide $[a,b]$ into $P_i$ intervals where $f$ would be monotonic, and then $\mathcal{W}(f,P_i)$ would be equal to the variation of $f$ on each of those $P_i$ intervals, making $\mathcal{W}(f,P) = V_f[a, b]$ in the worst case scenario.

Answer 1

Yes. Since $f$ is continuous there exist points $y_k,z_k$ in $[x_{k-1}, x_k]$ such that $M_k=f(y_k)$ and $m_k =f(z_k)$. So $\mathcal W(f,P) =\sum |f(y_k) -f(z_k)| \leq V_f[a,b]$.