How Do You Calculate the Total Variation of the Sign Function Over an Interval?

[brian_m.]

Hello,

I want to calculate the total variation $\left \| f \right \|_{V(\Omega)}$ with $\Omega=(-1,1)$ and $f(x)=\mathrm{sgn}(x)$.

The total variation of a function is defined as follows:

$\left \| f \right \|_{V(\Omega)} :=\sup\left \{ \int_\Omega f\ \mathrm{div} (v)\ dx \ | \ v \in C^1_c(\Omega)^n \text{ with } \left \| v \right \|_{\infty,\Omega}\leq 1 \right \}$

So, this is a very abstract definition and I don't know how to apply it...

Can you please help me?

Thank you in advance!

Bye,
Brian

 

[mighty2000]

Hello Brian,

The total variation of a function is a measure of the "roughness" or "variation" of the function over a given domain. In your case, the function f(x) is the sign function, which takes on values of -1 for x < 0 and 1 for x > 0. Since this function is not continuous, it may seem difficult to calculate its total variation using the given definition.

However, we can still apply the definition by considering a sequence of smooth functions v_n that converge to f(x) as n approaches infinity. For example, we could consider the sequence v_n(x) = x^n, which approaches the sign function as n approaches infinity. Then, we can use the definition to calculate the total variation of each v_n and take the supremum over all of these values to get the total variation of f(x).

In this case, we have:

\left \| f \right \|_{V(\Omega)} = \sup\left \{ \int_\Omega x^n\ \mathrm{div} (v)\ dx \ | \ v \in C^1_c(\Omega)^n \text{ with } \left \| v \right \|_{\infty,\Omega}\leq 1 \right \}

We can then use integration by parts to calculate the integral:

\int_\Omega x^n\ \mathrm{div} (v)\ dx = \int_{-1}^1 x^n\frac{\partial v}{\partial x}\ dx = \frac{1}{n+1} \left [x^{n+1}v\right ]_{-1}^1 - \frac{1}{n+1} \int_{-1}^1 x^{n+1}\frac{\partial v}{\partial x}\ dx

Since v is a smooth function with compact support, its derivative will also have compact support and we can apply the fundamental theorem of calculus to the first term to get:

\frac{1}{n+1} \left [x^{n+1}v\right ]_{-1}^1 = \frac{2}{n+1}

As n approaches infinity, this term goes to 0 and we are left with:

\left \| f \right \|_{V(\Omega)} = \sup\left \{ - \frac{1}{n+1} \int_{-1}^