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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?

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  • $\begingroup$ I saw your question yesterday, and it motivated me to think of this question, I have no answer, good question anyway. $\endgroup$ – user284331 Dec 7 '19 at 20:31
  • $\begingroup$ @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. $\endgroup$ – 0xbadf00d Dec 7 '19 at 20:34
  • $\begingroup$ @user284331 Regarding the ordering: It's the same as the one given on a general Hilbert lattice. $\endgroup$ – 0xbadf00d Dec 7 '19 at 20:38
  • $\begingroup$ 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? $\endgroup$ – Rhys Steele Dec 7 '19 at 21:23
  • $\begingroup$ @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 $c_1=c_2$. $\endgroup$ – 0xbadf00d Dec 8 '19 at 5:29
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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<Af,f\right>_{r}, \end{align*} and that \begin{align*} c_{3}=\sup_{f,g\geq 0,~\|f\|=\|g\|=1}\left|\left<Af,g\right>\right|, \end{align*} where \begin{align*} \left<f,g\right>=\int f\overline{g}, \end{align*} and that \begin{align*} \left<f,g\right>_{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 $\left<Af,f\right>_{r}=\left<Af,f\right>$ and $\left|\left<Af,g\right>\right|=\left<Af,g\right>\geq 0$, 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<Af,g\right>=\left<A(f+g),f+g\right>-\left<A(f-g),f-g\right>. \end{align*} But in this case, the term $\text{Re}\left<Af,g\right>$ simply becomes $\left<Af,g\right>$.

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

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

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