# An Integral inequality

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

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We are not mathematical-problem-solving-devices. So, my first advice would be to be more polite : "prove that..." is an order, and I won't have it. This said, where does this question comes from? Is this homework? What have you tried? – D. Thomine Mar 21 '12 at 17:01

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: $$|f({1 \over 2})| \leq \int_0^1|f(x)|\,dx+ {1 \over 2}\int_0^1 |f'(x)|\,dx$$

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We need this general statement for a real continuously differentiable function :

$$\int_a^b{f(t)dt} = bf(b) - af(a) - \int_a^b{tf'(t)dt}= (b-a)f(a) - \int_a^b{(t-b)f'(t)dt}$$

It is just an integration by part, but using two different primitives

Put $a = 0$ and $b=\frac{1}{2}$ in the first equality to get

$$\int_0^{\frac{1}{2}}{f(t)dt} = \frac{1}{2}f({\frac{1}{2}}) - \int_0^{\frac{1}{2}}{tf'(t)dt}$$

and $a = {\frac{1}{2}}$ and $b=1$ in the second equality

$$\int_{\frac{1}{2}}^1{f(t)dt} = {\frac{1}{2}}f({\frac{1}{2}}) - \int_{\frac{1}{2}}^1{(t-1)f'(t)dt}$$

And now it's easy : just take the sum of those two equalities and use triangle inequality, it will give you what you need

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