It is true for a continuous differentiable bounded function $f(x)$ that its total variation $V_a^b(f)\ge\max_{aI am trying to understand the intuition about the total variation: at first I believe it leads to the length of the codomain/range of the function $f(x)$, which I think is equivalent to $\Delta_y(f) = \max\{f(x)\} - \min\{f(x)\}$, but I didn't find any formal definition, or the formal name of this "codomain/range length", and also I understand now that the total variation have much more deep properties, as $f(x) - f(a) \leq V^b_a(f) \text{ for all } x \in [a , b ]$ , as example.
So I want to understand if the follow related statements are true for one-variable bounded continuous differentiable functions $f(x)$ (please note that I am using $f(x)$ and also its absolute value $|f(x)|$):

*

*There is a formal name for the "codomain/range length" $\Delta_y(f) = \max\{f(x)\} - \min\{f(x)\}$?? Is well defined?? (maybe I miss some absolute values $|\cdot|$).

*It is true $V^b_a(f) \geq \max_{a \leq x \leq b}\{f(x)\} - \min_{a \leq x \leq b}\{f(x)\}$???

*It is true $2 \max_{a \leq x \leq b}\{|f(x)|\}\geq \max_{a \leq x \leq b}\{f(x)\} - \min_{a \leq x \leq b}\{f(x)\}$???

*There is a way to relate the total variation $V^b_a(f)$ with  $\max_{a \leq x \leq b}\{|f(x)|\}$??? Specifically, is any of this two relations true: (a) $V^b_a(f) \geq 2\max_{a \leq x \leq b}\{|f(x)|\}$ or (b) $V^b_a(f) \leq 2\max_{a \leq x \leq b}\{|f(x)|\}$??

I don't have a deep mathematical background, so please try to make more intuitive than formal answers, as example, if you make a topological explanation it will be unintelligible for me. Beforehand, thanks you very much.
 A: 
"There is a formal name for the "codomain/range length" $\Delta_y(f) = \max\{f(x)\} - \min\{f(x)\}$?? Is well defined?? (maybe I miss some absolute values $|\cdot|$).

You could say, the diameter of the range of $f$, or $\operatorname{diam} \operatorname{Range} f$. The notation $\Delta_y(f)$ would also work so long as you define it beforehand. I don't know whether or not there is another term for it.
I don't think you're missing absolute values here, though the concept you're trying to describe is coming from you, not anyone else. Since $f$ is continuous, its range will be an interval, and what you wrote definitely describes the length of that interval.

It is true $V^b_a(f) \geq \max_{a \leq x \leq b}\{f(x)\} - \min_{a \leq x \leq b}\{f(x)\}$???

Yes, indeed this is true. The total variation measures the total vertical distance you travel while moving along the graph. You must travel at least from the maximum of the range to the minimum of the range. If $f$ isn't monotone increasing or decreasing, then you will have to travel more (i.e. the above is a strict inequality), retreading ground that you've already trodden.
As an example, consider the function $f(x) = \sin(x)$ over $[0, 2\pi]$. The range is $[-1, 1]$, and has diameter $2$. But, the total variation is $4$:

*

*First, $\sin$ increases from $0$ to $1$, adding $1$ to the total variation, then

*Second, $\sin$ decreases from $1$ to $-1$, adding $2$ to the total variation, and

*Finally, $\sin$ increases from $-1$ to $0$, adding $1$ to the total variation.

The decreasing bit in the middle retreads the range from $0$ to $1$, then covers the range from $0$ to $-1$, which is once again retrodden by the final increasing bit. If the function is only increasing or only decreasing (i.e. "monotone"), then no piece of the range is retrodden, and we get equality!

Is it true $2 \max_{a \leq x \leq b}\{|f(x)|\}\geq \max_{a \leq x \leq b}\{f(x)\} - \min_{a \leq x \leq b}\{f(x)\}$???

Yes, indeed it is. We can write
\begin{align*}
\max_{a \leq x \leq b}f(x) - \min_{a \leq x \leq b}f(x) &= \max_{a \leq x \leq b}f(x) + \max_{a \leq x \leq b}(-f(x)) \\
&\le \max_{a \leq x \leq b}|f(x)| + \max_{a \leq x \leq b}|-f(x)| \\
&= 2\max_{a \leq x \leq b}|f(x)|.
\end{align*}

There is a way to relate the total variation $V^b_a(f)$ with  $\max_{a \leq x \leq b}\{|f(x)|\}$??? Specifically, is any of this two relations true: (a) $V^b_a(f) \geq 2\max_{a \leq x \leq b}\{|f(x)|\}$ or (b) $V^b_a(f) \leq 2\max_{a \leq x \leq b}\{|f(x)|\}$??

No, neither are true necessarily.
The total variation can be made as large as you want, even restricting the range to a small interval. For example, $\sin(nx)$ in $[0, 2\pi]$ has total variation $4n$, since the curve performs $n$ full oscillations in $[0, 2\pi]$, and as we calculated above, each oscillation contributes $4$ to the total variation. This means that we will never get $V_a^b(f) \le \underline{\hspace{20pt}}$, where the right hand side depends only on the maximum or minimum value of $f$ or $|f|$, as the left hand side can be made to go to infinity, while the right hand side is bounded.
We also don't have $V_a^b(f) \ge 2\max_{a \le x \le b} |f(x)|$ because, if $f$ is a constant function, e.g. $f(x) = 1$ for all $x$, then $V_a^b(f) = 0$, but $2\max_{a \le x \le b} |f(x)| = 2$.
