How Does Total Variation Relate to the Integral of the Absolute Derivative?

[fourier jr]

problem: Let f \in BV[a,b]. Then \int_{a}^{b} |f'| \leq T_a^b f where T_a^b f is the total variation of f over [a,b].

there are some lemmas, etc that got me this far:
\int_{a}^{b} f' \leq f(b)-f(a) = P_a^b - N_a^b \leq P_a^b + N_a^b = T_a^b f where P is the positive variation & N is the negative variation of f.

the absolute value there messes me up; i don't know what to do about it. i know there's a theorem that says the following:
\vert \int_E f \vert \leq \int_E |f|
would that help at all?

 

[fourier jr]

up... can anyone help?

 

[wais]

Yes, the theorem that you mentioned can definitely help in this situation. By applying that theorem, we can rewrite the integral as \int_{a}^{b} |f'| = \int_{a}^{b} |f'| \cdot 1 \leq \int_{a}^{b} |f'| \cdot |f'| = \int_{a}^{b} |f'|^2.

Now, using the Cauchy-Schwarz inequality, we have \int_{a}^{b} |f'|^2 \leq \int_{a}^{b} |f|^2 \cdot \int_{a}^{b} |f'|^2.

Since f \in BV[a,b], we know that f' \in L^2[a,b], meaning that \int_{a}^{b} |f'|^2 is finite. Therefore, we can conclude that \int_{a}^{b} |f'| is also finite.

Combining this with the previous inequality, we get \int_{a}^{b} |f'| \leq T_{a}^{b} f, as desired. This shows that the total variation problem holds for functions in BV[a,b].