# Alternative definitions of variation of a function

Suppose $f \in L^1(\mathbb(R))$. Suppose $\int (\phi)(z):=\int_{x \leq z} \phi(x) dm(x)$.

I am trying to understand if/why the following are equivalent:

1) $\sup |\int f \phi \ dm|$, where sup is taken over all $\phi \in L^1$ with $\|\int(\phi)\|_\infty \leq 1$ and $\int \phi dm =0$. (G. Keller, Stochastic stability in some chaotic dynamical systems, Mh. Math 94, p.323)

2) $\sup \int f g' dm$, where $g$ is $C^1(\mathbb{R})$ with compact support and $|g| \leq 1$.(E. Giusti, Minimal surfaces and function of bounded variation, Birkh\"auser)

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You can easily check that the derivatives $g'$ of the functions $g$ that enter in the supremum in 2) are exactly the continuous $\phi$ used in the supremum 1). Moreover, the continuous $\phi$ are $L^1$ dense in all the $\phi$'s. Therefore the suprema are the same. –  Pietro Majer Jul 4 '11 at 7:52
note that this site is mainly devoted to research questions; please have a glance to the FAQ, where more suitable sites are also suggested. –  Pietro Majer Jul 4 '11 at 7:54
Thank you for the simple answer. For my future posts I will keep your second comment in mind. :) –  Banach Jul 4 '11 at 13:17
Could someone please explain what is wrong with the following? Consider the function $f=0$ on $[0,1/2]$, $f=1$ on $[1/2,1]$. Obviously, $V(f)=1$. But applying the above definition we have: $$V(f)=\sup \{\int_{1/2}^1 g'(x) dx \}= \sup \{g(1)-g(1/2) \} \text{.}$$ Now take $g$ such that $g(1/2)=-1$ and $g(1)=1$ (here $\|g\|_\infty \leq 1$). It follows that $V(f) \geq 2$. –  Banach Jul 4 '11 at 14:21
"Obviously $V(f) = 1$" are you sure? If you go up 1 and come back down 1, you traveled 2. –  Willie Wong Jul 4 '11 at 14:53