# Prove or disprove this argument

Let $L>0$ and let $\Omega$ be the set of all integrable functions from $[0,L]$ to $]0,+\infty[$.

For all $\varphi, \psi \in \Omega$ define $\left \langle \varphi,\psi \right \rangle:=\int_{0}^{L}\varphi(x)\psi(x)dx$.

Let $f,h\in \Omega$ such that $\left \langle f,h \right \rangle=\frac{1}{2}L^{2}$. Also, consider the set
$\omega(k):=\{g\in \Omega\colon \left \langle \textbf 1,g \right \rangle=1\wedge\left \langle h,g \right \rangle=k>0\}$,
where $\textbf 1\in \Omega$ is the function that maps everything to $1$. Note that $f$ or $h$ are not $\textbf 1$.

Is it true that for every $k\in]0,+\infty[$:

$\displaystyle \frac{\max \limits_{\large g\in \omega(k)}\left \langle f,g \right \rangle}{\min \limits_{\large g\in \omega(k)}\left \langle f,g \right \rangle}$ is constant, that is, does there exist $\alpha \in \Bbb R$ such that $\displaystyle \frac{\max \limits_{\large g\in \omega(k)}\left \langle f,g \right \rangle}{\min \limits_{\large g\in \omega(k)}\left \langle f,g \right \rangle}=\alpha$.

• what does "cte" means? – ougao Aug 18 '13 at 14:23
• It means: Constant. – Amir Kazemi Aug 18 '13 at 14:23
• @Amir $\huge\color{blue}{+}$ for your nice question ;) – Software Aug 18 '13 at 14:44
• I've completly rewritten the question to make it more clear. Hopefully everything is intelligible now and the OP can confirm that I didn't alter the meaning of the question. – Git Gud Aug 18 '13 at 16:06
• It should be "Let $\Omega$ be a set of" instead of "the set of all". That much is clear :) – fedja Aug 18 '13 at 16:30

Set $\Omega_1 := \{g \in \Omega \mid \langle \textbf 1,g\rangle = 1\}$. Then, is it true that $\displaystyle \frac{\max \limits_{\large g\in \Omega_1}\left \langle f,g \right \rangle}{\min \limits_{\large g\in \Omega_1}\left \langle f,g \right \rangle}$ is constant, that is, that there exists $\alpha \in \Bbb R$ such that $\displaystyle \frac{\max \limits_{\large g\in \Omega}\left \langle f,g \right \rangle}{\min \limits_{\large g\in \Omega}\left \langle f,g \right \rangle}=\alpha$?
In this case, the answer is no. For example, let $$f_1(x) = 1$$ and $$f_2(x) = \begin{cases} 1,\quad &x \in [0,1]; \\ 2,\quad &x \in ]1,2].\end{cases}$$ Then $$\frac{\max \limits_{\large g\in \Omega_1}\left \langle f_1,g \right \rangle}{\min \limits_{\large g\in \Omega_1}\left \langle f_1,g \right \rangle}=1,$$ but note that $\max_{g \in \Omega_1} \langle f_2,g\rangle \geq \frac{\|f_2\|_2^2}{\|f_2\|_1} = \frac53$, and $\min_{g \in \Omega_1} \langle f_2,g\rangle \leq 1$, (using functions $g$ which are close to being $1$ on $[0,1]$ and $0$ on $]1,2]$) so $$\frac{\max \limits_{\large g\in \Omega_1}\left \langle f_2,g \right \rangle}{\min \limits_{\large g\in \Omega_1}\left \langle f_2,g \right \rangle}\geq \frac53 > 1.$$
• $f$ is defined. $g$ is a variable! For a definite $f$, alter $g$ to find the max and min. Is it constant for every $g$? – Amir Kazemi Aug 18 '13 at 16:56
• @Aaron It should be 'that is, does there exist $\alpha \in \Bbb R\ldots$'. Same in the question, if you wanna correct it. – Git Gud Aug 18 '13 at 16:57