# Show $\sup_{0≤f∈L^2}\frac{\|Af\|_{L^2}}{\|f\|_{L^2}}=\sup_{0≤f∈L^2}\frac{⟨Af,f⟩_{L^2}}{\|f\|_{L^2}^2}$ for self-adjoint nonnegativity-preserving $A$

Let $$(E,\mathcal E,\mu)$$ be a measure space and $$A$$ be a self-adjoint bounded linear operator on $$L^2(\mu)$$. Assume $$Af\ge0$$ for all $$f\in\mathcal L^2(\mu)$$ with $$f\ge0$$. Let $$c_1:=\sup_{\substack{f\in\mathcal L^2(\mu)\setminus\{0\}\\f\ge0}}\frac{\left\|Af\right\|_{L^2(\mu)}}{\left\|f\right\|_{L^2(\mu)}}.$$

I would like to show that $$c_1=\sup_{\substack{f\in\mathcal L^2(\mu)\setminus\{0\}\\f\ge0}}\frac{\langle Af,f\rangle_{L^2(\mu)}}{\left\|f\right\|_{L^2(\mu)}^2}=:c_2.\tag1$$

Note that this is a classical result which is true when the suprema in the definitions of $$c_1$$ and $$c_2$$ are taken over $$\mathcal L^2(\mu)\setminus\{0\}$$ and we take the absolute value of the inner product in the definition of $$c_2$$.

Mimicing the usual proof, we easily obtain $$c_2\le c_1=\sup_{\substack{f,\:g\:\in\:\mathcal L^2(\mu)\\\left\|f\right\|_{L^2(\mu)}\:=\:\left\|g\right\|_{L^2(\mu)}\:=\:1\\f,\:g\:\ge\:0}}\langle Af,g\rangle_{L^2(\mu)}=:c_3\tag2.$$ Now we the classical claim (without the nonnegativity condition in the domain over which the supremum is taken in the definition of $$c_3$$) is concluded by showing $$c_3\le c_2\tag3$$ using the parallelogram law. Obviously, the problem with this is that the difference of nonnegative functions doesn't need to be nonnegative. Can we fix this or conclude by a different argument?

• I saw your question yesterday, and it motivated me to think of this question, I have no answer, good question anyway. – user284331 Dec 7 '19 at 20:31
• @user284331 If $\kappa$ is any transition kernel, then $\kappa f\ge0$ whenever $f\ge0$. Simply cause $(\kappa f)(x)=(\kappa(x,\;\cdot\;)f$ is the integral of a nonnegative function with respect to a nonnegative measure. In your situation, $K$ could be the density of a transition kernel with respect to the Lebesgue measure. – 0xbadf00d Dec 7 '19 at 20:34
• @user284331 Regarding the ordering: It's the same as the one given on a general Hilbert lattice. – 0xbadf00d Dec 7 '19 at 20:38
• I suspect that there is something wrong with the question as stated. Unless I misunderstand something, if you know $(2)$ then the thing you'd want to prove is that $c_1 \leq c_3$. To do this, it'd be enough to show that for $f \geq 0$ with $\|f\| = 1$ we have that $\|Af\| \leq c_3$. If $Af = 0$ this is trivial. Otherwise, we can take $g = \|Af\|^{-1} Af$ to see that $\|Af\| \leq \langle Af, g \rangle \leq c_3$. Maybe the $\sup$ in $c_3$ is meant to be taken over non-negative functions or something? – Rhys Steele Dec 7 '19 at 21:23
• @RhysSteele Maybe you were confused since I wrote $c_1\ge c_2$ (instead of $c_2\le c_1$); otherwise I don't know what you mean. If $(2)$ holds, then $c_2\le c_1=c_3$. So, if we can show that $c_3\le c_2$, we need to have $c_1=c_2$. – 0xbadf00d Dec 8 '19 at 5:29

I assume that the complex Hilbert space $$L^{2}(\mu)$$ is in issue, so the definition of $$c_{2}$$ and $$c_{3}$$ are respectively \begin{align*} c_{2}=\sup_{f\geq 0,~\|f\|=1}\left_{r}, \end{align*} and that \begin{align*} c_{3}=\sup_{f,g\geq 0,~\|f\|=\|g\|=1}\left|\left\right|, \end{align*} where \begin{align*} \left=\int f\overline{g}, \end{align*} and that \begin{align*} \left_{r}=\int fg. \end{align*}

Let $$f,g\geq 0$$, $$\|f\|=\|g\|=1$$ be given. First of all, since $$A(f)\geq 0$$, it is trivial that $$\left_{r}=\left$$ and $$\left|\left\right|=\left\geq 0$$, so there is no need to distinguish $$\left<\cdot,\cdot\right>_{r}$$ and $$\left<\cdot,\cdot\right>$$ in the definition of $$c_{2}$$ and $$c_{3}$$. The absolute value in $$c_{3}$$ can also be removed.

We know the formula that \begin{align*} 4\text{Re}\left=\left-\left. \end{align*} But in this case, the term $$\text{Re}\left$$ simply becomes $$\left$$.

Now we let $$h=f-g$$, the crucial point is to realize that \begin{align*} |A(h)|\leq A(|h|). \end{align*} Indeed, since $$h$$ is real-valued, we have $$|h|+h\geq 0$$. Since $$h=h^{+}-h^{-}$$, linearity of $$A$$ gives $$Ah=Ah^{+}-Ah^{-}$$. Keep in mind that both $$Ah^{+},Ah^{-}\geq 0$$, so $$Ah$$ is real-valued.

As $$A$$ is order-preserving, we have $$A(|h|+h)\geq 0$$, linearity of $$A$$ and the fact that $$Ah$$ being real-valued give $$A(|h|)\geq-A(h)$$. The same account applies to $$|h|-h\geq 0$$ to get $$A(|h|)\geq A(h)$$.

As a result, \begin{align*} \left|\left\right|\leq\left<|A(f-g)|,|f-g|\right>\leq\left. \end{align*} We obtain that \begin{align*} & 4\left\\ &\leq\|f+g\|^{2}\left|\left\right|+\|f-g\|^{2}\left|\left\right|\\ &=\|f+g\|^{2}\left+\|f-g\|^{2}\left\\ &\leq c_{2}(\|f+g\|^{2}+\|f-g\|^{2})\\ &= 2c_{2}(\|f\|^{2}+\|g\|^{2})\\ &= 4c_{2}, \end{align*} so $$\left\leq c_{2}$$, and hence $$c_{3}\leq c_{2}$$ is claimed.

• Thank you! What you wrote is indeed the crucial point. The rest is the same as for the standard result. – 0xbadf00d Dec 10 '19 at 15:31
• Note that the first equality after "We obtain that" should be "$\le$". – 0xbadf00d Dec 10 '19 at 15:31