Let $1≤p≤\infty$ and $ f \in L^{p}(a,b)$ such as there is a function $g \in L^{p}(a,b)$ that for all $\phi \in C^{1}(a,b)$ (and continuos in $[a,b]$) with $\phi(a)=\phi(b)=0$ we have:
$\int_{a}^{b} f(x)\phi'(x)dx= - \int_{a}^{b}g(x)\phi(x)dx.$
Prove that $f$ is absolute continuous in $[a,b]$ and there are constants $c>0$ and $0<\alpha<1$ such as for all $x, y \in [a,b]$
$|f(x)-f(y)|\leq c|x-y|^\alpha$
I really don't know where to start. If I can prove that $f(x)=f(a)+\int_{a}^{x} g(t) dt$ for all $x \in [a,b]$ then f would be absolute continuous, but I don't know how to prove that. The second part, well it seems really hard. I really appreciate your help. Thank you!
If you want to prove that $f(x) = f(a) + \int_a^x g(t) dt$, then that's equivalent to showing that $f(x) - f(a) - \int_a^x g(t) dt = 0$ (by $=$, I mean almost everywhere); to this end, notice that
\begin{align*} \int_a^b \Big(f(x) - f(a) - \int_a^x g(t) dt\Big) \phi'(x) dx &= -\int_a^b g(x) \phi(x) dx - f(a) \int_a^b \phi'(x) dx \\&\quad- \int_a^b \int_a^x g(t) dt \phi'(x) dx \end{align*}
The middle integral vanishes because $\phi$ is zero on either endpoint. The last integral can be computed by changing variables:
\begin{align*} \int_a^b \int_a^x g(t) dt \phi'(x) dx &= \int_a^b \int_t^b \phi'(x) g(t) dx dt \\ &= \int_a^b \Big(\phi(b) - \phi(t)\Big) g(t) dt \\ &= - \int_a^b \phi(t) g(t) dt \end{align*}
and the terms cancel. You can then work out how this implies that the given expression is actually zero almost everywhere.
For the other part, now that we know $f$ is absolutely continuous, we can write $(y > x)$
$$f(y) - f(x) = \int_x^y g(t) dt$$
By the triangle inequality, for $p \ne 1, \infty$,
$$|f(y) - f(x)| \le \int_x^y |g(t)| dt \le \left(\int_x^y |g|^p\right)^{1/p} \left(\int_x^y 1^{p'}\right)^{1/p'} \le \|g\|_p |x - y|^{1/p'}$$
by Holder's inequality, where $p$ and $p'$ are conjugate exponents.
Can you finish from here?