# Trig Integral Inequality

Show that if $$f$$ is Riemann integrable on $$[a,b]$$, then

$$\left(\int_{a}^{b}f(x)\sin x\ dx\right)^2+\left(\int_{a}^{b}f(x)\cos x\ dx\right)^2\le(b-a)\int_{a}^{b}f^2(x)\ dx.$$

I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $$\left(\int_{a}^{b}\sin^2x+\cos^2x\ dx\right)\left(\int_{a}^{b}f^2(x)\ dx\right)$$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!

• What's $\sin(x)^{2}+\cos(x)^{2}$ equal to? Jan 21, 2019 at 16:13
• The direction of the inequality should be reversed. It almost immdiately follows from Caychy-Schwarz. Jan 21, 2019 at 16:14

By C-S $$(b-a)\int\limits_a^bf(x)^2dx=\int\limits_a^bf(x)^2\sin^2xdx\int_a^b1dx+\int\limits_a^bf(x)^2\cos^2xdx\int_a^b1dx\geq$$ $$\geq\left(\int\limits_a^bf(x)\sin{x}dx\right)^2+\left(\int\limits_a^bf(x)\cos{x}dx\right)^2.$$ Actually, $$f^2(x)=f(f(x))$$ and $$f(x)^2=\left(f(x)\right)^2.$$

1.Prove that $$:=\int_a^bf(x)g(x)dx$$ is an inner product.

2.Observe that the norm it induces is $$\|f\|=\big{(}\int_a^bf^2(x)dx\big{)}^{1/2}$$.

3.Rewrite your inequality as $$^2+^2\leq(b-a)\cdot\|f\|^2$$

4.Using Cauchy-Schwarz, show that $$^2+^2 \leq \|f\|^2(\|\sin\|^2+\|\cos\|^2)$$

5.Observe that $$\|\sin\|^2+\|\cos\|^2=\int_a^b\sin(x)^2dx+\int_a^b\cos(x)^2dx=\int_a^b1dx=b-a$$.

$$\left(\int_{a}^{b}f(x)\sin x\ dx\right)^2+\left(\int_{a}^{b}f(x)\cos x\ dx\right)^2=\Big|\int^b_a e^{ix} f(x)\,dx\Big|^2$$

An application of Cauchy-Bunyakovsky's inequality yields $$\Big|\int^b_a e^{ix} f(x)\,dx\Big|\leq\Big(\int^b_a|e^{2ix}|^2\,dx\Big)^{1/2}\Big(\int^b_a|f(x)|^2\,dx\Big)^{1/2}=(b-a)^{1/2}\int^b_a|f(x)|^2\,dx$$