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The question is to show that, given $f$ is a non-negative Riemann integrable function over an interval $[a,b]$, such that $\int_a^b f=1$, then we have that $$\left(\int_a^b f(x) \cos(x)\right)^2+\left(\int_a^b f(x) \sin(x)\right)^2\leq 1$$ And I cannot think of the correct trick here.

With the Cauchy-Schwartz inequality, I have gotten that $$\left(\int f \cos\right)^2+\left(\int f \sin\right)^2\leq \int f^2\int \cos^2+\int f^2\int \sin^2=\int f^2\int 1=(b-a)\int f^2$$ Which gives me almost the inequality I want, but how can I then argue about the value of $\int f^2$ working out properly? Or is my approach going completely up the wrong tree?

Brandon Myers
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  • What happens if you write $$\biggl( \int_a^b f(x)\cos x,dx \biggr)^2 = \int_a^b \int_a^b f(x)f(y)\cos x\cos y,dx,dy$$and similarly for the other integral? – Greg Martin Feb 28 '23 at 19:59
  • @Greg Martin I don't see how that helps. Writing it as a double integral allows you to add the two together, but then you need to get to an inequality somewhere, and also now have to deal with the different variables of integration. (Additionally, I am not 100% sure of that expression?) – Brandon Myers Feb 28 '23 at 20:39
  • I think we have $\cos x\cos y+\sin x\sin y \le (\cos^2x+\sin^2x)^{1/2}(\cos^2y+\sin^2y)^{1/2}$ by Cauchy–Schwarz – Greg Martin Feb 28 '23 at 23:21

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It might be a bit overkill, but if you know a bit about Lebesgue theory, using Jensen's inequality on $f(x) \mathrm dx$ (since $\int f = 1$ and $f \geq 0$) you can conclude. I guess that this version of Jensen's inequality remains accessible when only knowing about Riemann integration :

Let $f, g$ be two continuous real functions defined on $(a, b)$ such that $\int_a^b f(x) \mathrm d x = 1$ and $f$ is non-negative, and $\varphi$ a continuous convex function defined over $\mathbb R$. Then, $$ \varphi\left(\int_a^b g(x) f(x) \mathrm d x\right) \leq \int_a^b \varphi \circ g(x) f(x) \mathrm d x.$$

Here, where $\varphi$ is the square function and $g = \cos$ or $\sin$,

$$\left(\int_a^b \cos x f(x) \mathrm d x \right)^2 \leq \int_a^b \cos^2 x f(x) \mathrm d x \quad \textrm{and} \quad \left(\int_a^b \sin x f(x) \mathrm d x \right)^2 \leq \int_a^b \sin^2 x f(x) \mathrm d x. $$

Then,

$$\left(\int_a^b \cos x f(x) \mathrm d x \right)^2 + \left(\int_a^b \cos x f(x) \mathrm d x \right)^2 \leq \int_a^b (\cos^2 x + \sin^2 x) f(x) \mathrm d x = 1.$$

Zag
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  • This just kicks the can down the road to proving Jensen's Inequality, which, given the class hasn't talked about convex functions at all, seems like it is not remotely intended. – Brandon Myers Feb 28 '23 at 20:45