Suppose that $f \in \mathcal{C}^{1}([a, b])$. Prove that $|f(x)| \leqslant \frac{1}{2}|f(a)+f(b)|+\int_{a}^{b}\left|f^{\prime}\right|$

Question

I've tried to prove it through the following way but failed. There is a gap that I can't get through.

$$ |f(x)| = \left|\int^b_x f'-f(b)\right| = \left| \int^x_a f'+f(a)\right|, $$ So $$ 2|f(x)|=\left|\int^b_x f'-f(b)\right| + \left|\int^x_a f'+f(a)\right| $$ Thus $$ 2|f(x)| = \left|\int^b_x f'-f(b)\right| + \left|\int^x_a f'+f(a) \right| \le \left| \int^b_x f'\right| + |f(b)| + \left|\int^x_a f'\right|+|f(a)| $$ Since we have $$ \left| \int^b_x f' \right| \le \int^b_x |f'| \quad \text{ and } \quad \left| \int^x_a f' \right| \le \int^x_a |f'| $$ Thus $$ \left|\int^b_x f'\right| + \left| \int^x_a f' \right| \le \int^b_x |f'| + \int^x_a |f'| = \int^b_a |f'| $$ But the problem is that I can't have $|f(b)|+|f(a)|\le|f(a)+f(b)|$ So I think I probably went in a wrong way.

Answer 1

I believe this is the trick you are missing: Since $f$ is continuous, there exists $c$ such that $f(c)=\frac{f(a)+f(b)}2$.

Now $|f(x)|\le |f(c)|+\int_c^x| f’|$ and it’s easy to finish.