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Integral and inequality

Let f:[0,1] \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} \int_0^1 f(x) \, \mathrm{d}x = \kappa =\int_0^1 x f(x) \, \mathrm{d}x \end{equation*}

Prove that

    \[\int_0^1 f^2(x) \, \mathrm{d}x \geq 4\kappa^2\]

Solution

We note that

    \begin{align*} \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x &= \int_{0}^{1} f^2(x) \, \mathrm{d}x -12 \kappa \int_{0}^{1} x f(x) \, \mathrm{d}x + 36 \kappa^2 \int_{0}^{1} x^2 \, \mathrm{d}x \\ &=\int_{0}^{1} f^2(x) \, \mathrm{d}x - 12 \kappa \int_{0}^{1} f(x) \, \mathrm{d}x + \frac{36\kappa^2}{3} \\ &=\int_{0}^{1} f^2 (x) \, \mathrm{d}x - 12 \kappa^2 + 12 \kappa^2\\ &= \int_{0}^{1} f^2(x) \, \mathrm{d}x \end{align*}

On the other hand by Cauchy – Schwartz we have that

    \begin{align*} \int_{0}^{1} f^2(x) \, \mathrm{d}x &= \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x \\ &\geq \left [ \int_{0}^{1} \left ( f(x) - 6 \kappa x \right ) \, \mathrm{d}x \right ]^2 \\ &=\left ( \kappa - 3 \kappa \right )^2 \\ &= 4\kappa^2 \end{align*}

The result follows.

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