Suppose $f$ is absolutely continuous, real valued, defined on $[a,b]$. Further we are given that $\int_{a}^{b}f = 0.$ If $1\leq p < \infty$, we are to show that $|f(x)| \leq 2 (b-a)^{1/p*}||f'||_p$.
I was able to show that $|\int_{a}^{t}f'(t)| \leq (b-a)^{1/p*}||f'||_p $, using Holder's inequality. Not sure how to proceed from here. How do I use that if $f $ is absolutely continuous then $f(x) = f(a) + \int_{a}^{t}f'(t)$?
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Define $F(x) = \int_{a}^{x}f(t)dt$. By fundamental theorem of calculus $F$ is differentiable and $F'(x)=f(x)\:\: \forall x \in (a,b)$.
Note that $F(a)=0=F(b)$. So by Rolle's theorem there's a $c \in (a,b)$ such that $F'(c)=0=f(c)$.
By this theorem, $$f(x)-f(a)=\int_{a}^{x}f'(t)dt \tag{1}\label{1}$$ Evaluating at $c$,$$-f(a)=\int_{a}^{c}f'(t)dt \tag{2}\label{2}$$ Equations (1) and (2) gives us the result $$\begin{align} |f(x)|&=| -f(a)+\int_{a}^{x}f'(t)dt|\\ &\leq |\int_{a}^{c}f'(t)dt|+|\int_{a}^{x}f'(t)dt|\\ &\leq 2\int_{a}^{b}|f'(t)|dt\\ &\leq 2(b-a)^{1/p^*}\lVert f'\rVert_p \end{align}$$ The last step is because of Hölder's inequality
If $f$ is absolutely continuous then, by a generalized fundamental theorem of calculus, $f$ has a derivative $f'$ almost everywhere which is integrable and
$$f(x) = f(a) + \int_a^x f'(t) \, dt$$
Thus, applying the triangle and Holder's inequalities, we have
$$\tag{*}|f(x)| \leqslant |f(a)| + \int_a^x |f'(t)| \, dt \leqslant |f(a)| + \int_a^b |f'(t)| \, dt \\ \leqslant |f(a)| + \|f' \|_p (b-a)^{1 - 1/p}$$
Note that
$$0 = \int_a^b f(x) \, dx = f(a)(b-a) + \int_a^b \int_a^x f'(t) \, dt \, dx,$$
and it follows that
$$\tag{**}\begin{align}|f(a)| &= (b-a)^{-1} \left|\int_a^b \int_a^x f'(t) \, dt \, dx\right| \\ &\leqslant (b-a)^{-1} \int_a^b \int_a^x |f'(t)| \, dt \, dx \\ &\leqslant (b-a)^{-1} \int_a^b \int_a^b |f'(t)| \, dt \, dx \\ &\leqslant (b-a)^{-1} \int_a^b\|f' \|_p (b-a)^{1 - 1/p} \, dx \\ &= (b-a)^{-1} (b-a)\|f' \|_p (b-a)^{1 - 1/p} \\ &= \|f' \|_p (b-a)^{1 - 1/p}\end{align} $$
Using (*) and (**) we get
$$|f(x)| \leqslant 2\|f' \|_p (b-a)^{1 - 1/p}$$