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Here $\lvert \lvert P \rvert \rvert_\infty$ is the maximum absolute row sum of the matrix $P$, which is $1$ when $P$ is row stochastic. While the hypothesis in the title seems to hold when tested, I have not been able to improve on the trivial bound \begin{eqnarray*} \left| \left| (2Q-I)(2P-I)\right| \right|_\infty &\leq &\lvert \lvert 2Q - I \rvert \rvert_\infty \lvert \lvert 2P-I \rvert \rvert_\infty \\ &\leq & \left(2\lvert \lvert Q \rvert \rvert_\infty+\lvert \lvert I \rvert \rvert_\infty \right)\left(\lvert \lvert P \rvert \rvert_\infty+2\lvert \lvert I \rvert \rvert_\infty \right) \\ & = & 3 \cdot 3 =9, \end{eqnarray*} and anything better would be useful. The assumptions in the title imply that the matrix $(2Q-I)(2P-I)$ has rows summing to one and eigenvalues in $[-1,1]$.

The claim clearly holds for scalars. In two dimensions, define \begin{eqnarray} Q=\left( \begin{array}{cc} q_1 & 1- q_1 \\ 1-q_2 & q_2 \end{array} \right) \qquad P=\left( \begin{array}{cc} p_1 & 1- p_1 \\ 1-p_2 & p_2 \end{array} \right), \end{eqnarray} and since $Q$ is stochastic, it has the eigenvalue $1$. The other eigenvalue is $1-q_1-q_2$ because the trace equals the sum of the eigenvalues. Requiring the eigenvalues to be nonnegative then implies $q_1+q_2\geq 1$ and similarly for $P$.

The bound is achieved for $q_1=0$, $q_2=1$, $p_1=\frac{1}{2}$ $p_2=\frac{1}{2}$, as we have \begin{eqnarray} \left(2Q-I\right)\left(2P-I\right) &=& \left(\begin{array}{rr} -1 & 2 \\ 0 &1 \end{array} \right) \left(\begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}\right) =\left( \begin{array}{rr} 2 & -1 \\ 1 & 0 \end{array} \right), \end{eqnarray} where the maximum absolute row sum is 3. In the above we may also confirm that increasing $q_1$, $p_1$ or $p_2$ marginally reduces the $\infty$-norm, while reducing $p_1$ or $p_2$ increases the $\infty$-norm beyond 3 but violates the assumption that the eigenvalues of $P$ are nonnegative.

Any help is appreciated. The purpose of this is to bound the Jacobian of a particular map.

EDIT: A stronger hypothesis is $\lvert \lvert (2Q-I)(2P-I) \rvert \rvert_\infty\leq \max \left( \lvert \lvert (2Q-I) \rvert \rvert_\infty,\lvert \lvert (2P-I) \rvert \rvert_\infty \right)$

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Your conjecture is false for every $n\ge2$. Let $Q=ee_1^T$ and $P=ee_n^T$, where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb R^n$ and $e=e_1+\cdots+e_n$ is the all-one vector. Then \begin{aligned} (2Q-I)(2P-I)&=(2ee_1^T-I)(2ee_n^T-I)\\ &=4ee_n^T-2ee_1^T-2ee_n^T+I\\ &=2ee_n^T-2ee_1^T+I. \end{aligned} Therefore $\|(2Q-I)(2P-I)\|_\infty=5>3=\|2Q-I\|_\infty=\|2P-I\|_\infty$.

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  • $\begingroup$ Thank you for the counterexample. Do you think the 5 is an upper bound? Like I mentioned, any bound that's better than 9 would be very useful. It seems like your construction maximizes the norm. $\endgroup$ – Asco Sep 6 '19 at 16:41
  • $\begingroup$ @Asco I don't know. To be frank, I have no idea what a generic row stochastic matrix with nonnegative spectrum looks like. $\endgroup$ – user1551 Sep 6 '19 at 17:18

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