TITLE: An Integral inequality
QUESTION [5 upvotes]: Let $f$ be a continuously differentiable function on $[0,1]$. Prove that $$|f(\frac{1}{2})|\leq\int\limits_0^1|f(x)|dx+\frac{1}{2}\int\limits_0^1|f'(x)|dx$$

REPLY [3 votes]: Let $x_0 \in [0,1]$ be such that $|f(x_0)| \leq |f(x)|$ for all $x \in [0,{1 \over 2}]$. Then you have
$$f({1 \over 2}) = f(x_0) + \int_{x_0}^{1 \over 2} f'(x)\,dx$$
So
$$|f({1 \over 2})| \leq |f(x_0)| + \int_{x_0}^{1 \over 2} |f'(x)|\,dx$$
$|f(x_0)|$ is at most the average of $|f(x)|$ on $[0,{1 \over 2}]$, namely ${\displaystyle 2\int_0^{1 \over 2}|f(x)|\,dx}$, and also ${\displaystyle \int_{x_0}^{1 \over 2} |f'(x)|\,dx \leq \int_{0}^{1 \over 2} |f'(x)|\,dx} $. So the above gives
$$|f({1 \over 2})| \leq 2\int_0^{1 \over 2}|f(x)|\,dx+ \int_{0}^{1 \over 2} |f'(x)|\,dx$$
The corresponding inequality over the right half interval $[{1 \over 2}, 1]$ is 
$$|f({1 \over 2})| \leq 2\int_{1 \over 2}^1|f(x)|\,dx+ \int_{1 \over 2}^1 |f'(x)|\,dx$$
Adding the last two equations and dividing by two gives the desired inequality:
$$|f({1 \over 2})| \leq \int_0^1|f(x)|\,dx+ {1 \over 2}\int_0^1 |f'(x)|\,dx$$