# Prove an inequality based on a scalar product

Let $$V=\{f:[-\pi,\pi]\to\mathbb{R}| f\text{ continuous}\}$$ with:

$$f\times g=\int_{-\pi}^{\pi}f(x)g(x)dx$$

Show that $$\forall f \in V$$:

$$\left|\int_{-\pi}^{\pi}f(x)\sin(x)dx\right|\leq\sqrt{\pi}\sqrt{\int_{-\pi}^\pi f^2(x)dx}$$

I tried showing this with the abosolute value property of the integral:

$$|\int_a^bf(x)dx|\leq\int_a^b|f(x)|dx$$

but it didn't work out.. any ideas?

Hint:

Use the Cauchy–Bunyakovsky–Schwarz inequality.

• yup right, solved it, thanks! – C. Cristi Oct 25 '18 at 14:23