Prove these can't both be true: $\int_{0}^{\pi} [f(x) - \sin x] ^2 \, dx \le \frac{4}{9}$, $\int_{0}^{\pi} [f(x) - \cos x] ^2 \, dx \le \frac{1}{9}$

Question

If $f(x) \in L^2[0, \pi]$ Prove that these two inequalities cannot both be true:

$$ \int_{0}^{\pi} [f(x) - \sin x] ^2 \, dx \le \frac{4}{9} $$

$$ \int_{0}^{\pi} [f(x) - \cos x] ^2 \, dx \le \frac{1}{9} $$

I have been trying the following approach. Define two functions $G(x)$ and $H(x)$ such that $G(x) = f(x) - \sin x$ and $H(x) = \cos x - f(x)$. Then consider that $G+H$ in $L^2[0,\pi]$. I was adding the two functions together and using the triangle inequality of the norm but couldn't get the problem to go. Is this the right approach? Is there another direction for this problem?

Thank you.

Answer 1

This is the right approach. Let $u(x)=f(x) - \sin x$ and $w(x)=f(x) - \cos x$. The asummptions tell us that
$$ \lVert u\rVert\leq2/3, \quad \lVert w\rVert\leq1/3 $$ but $$ \lVert u-w\rVert=\sqrt{\pi} $$ as you can compute (or use a computer) where the norms are $L^2$ norms. Thus if both inequalities in the assumption were true then the triangle inequality would be violated, a contradiction. It must be the case that at least one of the inequalities in the assumptions is false.

Answer 2

Show that in this case

$$d(\sin,\cos)>d(\sin,f)+d(f,\cos)$$

where $d(f,g)=\| f -g \|_{L^2}$. This contradicts the triangle inequality.

Answer 3

Suppose they are both true.

By inequality between arithmetic mean and quadratic mean we have: $${1\over 2}(a+b)^2\leq a^2+b^2$$ we have

$${1\over 2} \int_{0}^{\pi} [(\sin x -f(x)) +(f(x)- \cos x)] ^2 \, dx \leq \int_{0}^{\pi} [f(x) - \sin x] ^2 \, dx + \int_{0}^{\pi} [f(x) - \cos x] ^2 \, dx\le \frac{5}{9} $$

so we have $$ \int_{0}^{\pi} [\sin x - \cos x] ^2 \, dx \le \frac{10}{9} $$

but $$ \int_{0}^{\pi} [\sin x - \cos x] ^2 \;dx = \int_{0}^{\pi} [1-\sin 2x] \;dx = \pi$$

A contradiction.