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I'm learning about functions of bounded variations and need to verify my work for this problem:

Show that $\| f \|_{BV} = | f(a) | + V_{a}^{b} f$ defines a norm in the space $BV[a,b]$.

My attempt and Thoughts:

First. We want to show that $\| f \|_{BV} \geq 0$ (the norm is always positive).

This is the same as writing that $| f(a) | + V_{a}^{b} f \geq 0$ (by the definition of $\| f \|_{BV}$).

Well, clearly $| f(a) | \geq 0$ and since $V_{a}^{b} f$ is defined as

$$V_{a}^{b} f = \sup_{P} V(f, P)$$ and we know that $V(f, P) \geq 0$. Hence, $\| f \|_{BV} \geq 0$.

Second. We want to show that $\| f \|_{BV} = 0 \iff f = 0$.

"$\impliedby$": if $f = 0$ i.e. $f(x) = 0, \forall{x} \in [a,b]$ then $\| f \|_{BV} = 0$.

"$\implies$": if $\| f \|_{BV} = 0$, then $| f(a) | + V_{a}^{b} f = 0$ if and only if $f$ is constant.

Third. We want to show that $\|\alpha \ f \|_{BV} = | \alpha | \| f \|_{BV}$.

We have $\|\alpha \ f \|_{BV} = |\alpha \ f(a) | + V_{a}^{b} (\alpha \ f) = | \alpha | | f(a) | + | \alpha | V_{a}^{b} f = | \alpha |(| f(a) | + V_{a}^{b} f) = | \alpha | \| f \|_{BV}$.

Fourth. We must prove that $\| f + g \|_{BV} \leq \| f \|_{BV} + \| g \|_{BV}$.

We have that $\| f + g \|_{BV} = | f(a) + g(a) | + V_{a}^{b} (f + g) \leq | f(a) | + |g(a) | + V_{a}^{b} f + V_{a}^{b} g = \| f \|_{BV} + \| g \|_{BV}$.


Since I'm new to this subject I would like to verify that my work is correct and know if there are parts of the demonstration that I can improve.

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    $\begingroup$ In second, you should state that $f=0$. The proof looks fine for me. $\endgroup$ – Hanul Jeon Dec 15 '15 at 10:12
  • $\begingroup$ The " $\implies$ " in the second step needs stating why $f=0$ (and not any other constant). The remaining seems correct. However, for the next time, a more detailed definition of $V_a^bf$ (what is $V(f,P)$ or why does it have the properties you are assuming?) would help other other readers in MSE follow your proof, even if it would not be so necessary in your work. $\endgroup$ – AugSB Apr 3 '16 at 10:16

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