# $|f(\frac{1}{2})|\leq \int_{0}^{1}|f(t)|dt+\frac{1}{2}\int_{0}^{1}|f'(t)|dt$

Let $f'\in\mathcal{R}[0,1]$ then prove the above inequality.

Try : What I have found till now is that $$\left|\frac{f(1)+f(0)}{2}\right|\leq \int_{0}^{1}|f(t)|dt+\frac{1}{2}\int_{0}^{1}|f'(t)|dt$$ and $$\left|\frac{f(1)}{2}\right|\leq \int_{0}^{1}\left|f(t)\right|dt+\frac{1}{2}\int_{0}^{1}|f'(t)|dt.$$ By using the integral $\int tf(t)dt$ and $\int (1-t)f(t)dt$. But I don't understand is that how do I get the $f(\frac{1}{2})$?

• Are you assuming $f$ is continuous? Otherwise we can redefine $f(1/2)$ to be arbitrarily large and this doesn't affect the integrals. – Luiz Cordeiro Dec 8 '16 at 18:21
• @LuizCordeiro well it's differentiable, so of course continuous. – user384138 Dec 8 '16 at 18:21
• @OpenBall I'm asking since the usual results hold for functions which are only differentiable a.e., or Lebesgue integrable, etc... It wasn't clear that this is the classical derivative. Anyway, from my previous comment we see that there should be some kind of approximation argument involved, maybe something similar to the first inequality you obtained. – Luiz Cordeiro Dec 8 '16 at 18:24
• Here $f'$ is riemann integrable and the primitive is $f$. I cant find any inequality that gives $f(1/2)$ – mudok Dec 8 '16 at 18:35

Your idea is essentially right, we just need to use different boundaries and integration by parts: $$\int_0^{1/2}tf'(t)dt=[tf(t)]^{1/2}_0-\int_0^{1/2}f(t)dt$$ so $$\frac{f(1/2)}{2}=\int_0^{1/2}f(t)dt+\int_0^{1/2}tf'(t)dt.\tag{1}$$ Now do the same analysis with $g(t)=f(1-t)$: $$\frac{f(1/2)}{2}=\frac{g(1/2)}{2}=\int_0^{1/2}g(t)dt+\int_0^{1/2}tg'(t)dt=\int_{1/2}^1f(t)dt+\int_{1/2}^1(1-t)f'(t)dt\tag{2}$$ Add (1) and (2): \begin{align*} |f(1/2)|&=|\int_0^1 f(t)dt+\int_0^{1/2}tf'(t)dt+\int_{1/2}^1(1-t)f'(t)dt|\\ &\leq\int_0^1|f(t)|dt+\int_0^{1/2}|t||f'(t)|dt+\int_{1/2}^1|1-t||f'(t)|dt\\ &\leq\int_0^1|f(t)|dt+\frac{1}{2}\int_0^1|f'(t)|dt \end{align*}