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Inequality of a function

Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative. Prove that:

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1} \left | f(x) \right | \, \mathrm{d}x + \frac{1}{2} \int_{0}^{1} \left | f'(x) \right | \, \mathrm{d}x\]

Solution

For 0 \leq x \leq \frac{1}{2} it holds that

    \[f\left ( \frac{1}{2} \right ) -f(x) = \int_{x}^{1/2} f'(t)\, \mathrm{d}t\]

Taking absolute values and using basic properties of the integral we get

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \left | f(x) \right | + \int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t\]

Integrating we have:

(1)   \begin{equation*}   \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1/2}\left | f(x) \right | + \frac{1}{2}\int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t  \end{equation*}

Working similarly on \left[\frac{1}{2} , 1 \right] we get

(2)   \begin{equation*} \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{1/2}^{1}\left | f(x) \right | + \frac{1}{2}\int_{1/2}^{1} \left | f'(t) \right | \, \mathrm{d}t    \end{equation*}

Adding equations (1), (2) we get the result.

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