# Reverse Riesz-Thorin inequality

We know that Riesz-Thorin inequality reads $||g||_{p_{\theta}}\leq||g||_{p_{0}}^{\theta}||g||_{p_{1}}^{1-\theta}$, where $0<p_0<p_1<+\infty$, $0<\theta<1$ and ${1\over{p_{\theta}}}={{\theta}\over{p_{0}}}+{{1-\theta}\over{p_{1}}}$. Let us further assume that $g$ is strictly positive and continuous on the domain $(0,1)$ (if necessary we can also assume that $0<m\leq g(x)\leq M<+\infty$ for all $x\in (0,1)$). My question is: does some kind of reverse Riez-Thorin inequality holds? By this I have in mind the reverse resembling, for instance, the relationship of the Holder inequality and the reverse Holder inequality. So, I guess, the reverse Riez-Thorin should read something like:

$||g||_{p_{\theta}}\geq||g||_{p_{0}}^{\theta}||g||_{p_{1}}^{1-\theta}$, where $0<p_0<1$, $p_1<0$, $0<\theta<1$ and ${1\over{p_{\theta}}}={{\theta}\over{p_{0}}}+{{1-\theta}\over{p_{1}}}$,

where, I believe, the natural choice for $p_{\theta}$ is $0<p_{\theta}<1$.

Is this true? I think this inequality may follow from the generalized reverse Holder inequality, which reads: $\int\Pi_{j=1}^m|g_j|\geq\Pi_{j=1}^{m}||g_j||_{r_j}$, where $\sum_{j=1}^{m}{1\over{r_j}}=1$, $0<r_1<1$ and $r_j<0$ for every $j=2,\dots,m$. So far, by choosing $m=2$ etc., I was able to prove the following:

If $0<\theta<1$ and ${1\over{r_1}}+{1\over{r_2}}={1\over{\theta^2}}$, where $0<r_1<1$ and $r_2<0$, then we have $||g||_{\theta}\geq||g||_{r_1}^{\theta}||g||_{r_{2}}^{1-\theta}$. But I do not know if this inequality matches exactly the reverse Riez-Thorin as stated above. Maybe such reverse is not true, I do not know (maybe it is more natural to choose $\theta>1$ as the interpolation parameter?). Any reference to papers or other literature is welcome. I tried to find sources on web, but they are scarse. Also, note that from the generalized reverse Holder inequality, by choosing $\theta:={1\over{m}}\in (0,1)$, $m\geq 2$, we get:

$||g||_{1\over{\theta}}\geq||g||_{r_1}^{\theta}||g||_{r_{2}}^{1-\theta}$, where $0<r_1<1$, $r_2<0$ and ${{\theta}\over{r_1}}+{{\theta-1}\over{r_2}}={\theta}$,

but I don't believe that this is a sharp inequality, since the norm on the left side of the inequality is the usual Lp norm for $p>1$. Still, one related question is whether the last inequality holds true for arbitrary $0<\theta<1$.