Holder's Inequality for $0

Question

I would like to ask about the constant $C$ for which the following inequality holds.

Let $0<p<1$ and let $q$ be its Hölder conjugate, that is $1/p+1/q=1$, then Is there $C>0$ such that $$||fg||_1 \leq C||f||_p||g||_q?$$

Answer 1

As Martin Argerami has pointed out in comments, no $C$ can make the inequality true. Martin gives the following counterexample: take $p=1/2,q=-1$, and consider the inequality for points $f=(1,0),g=(n, \frac{n}{n-1}) \in \mathbb R^2$. Then $\|f\|_p=\|g\|_q=1$, while $\|fg\|_1=n$. This also has the obvious generalisation to all such $p,q$ with $p\in(0,1)$, $q<0$, replacing $g$ with $g=((\frac1n)^{1/q} , (\frac{n-1}n)^{1/q})$.

In addition, the opposite inequality holds. The rest of the answer is to prove this:

Claim: For $0<p<1$, $f,g>0$ (just to avoid absolute values everywhere), $$ \int fg \ge \|f\|_{L^p} \|g\|_{L^q}.$$ where the definition of the "norms" are taken at face value: $$\|g\|_{L^q} := \frac1{(\int g^q)^{-1/q}}.$$

Proof: We need to prove, setting $r=-q>0$, $$ \|f\|_{L^p} \le \left(\int g^{-r}\right)^{1/r}\int fg. $$ we have $$ \int f^p = \int (fg)^p g^{-p}$$ this looks like it could be solved by an application of classical Holder's. We define $P$ by $Pp=1$ i.e. $P:=1/p$, and the Holder conjugate $Q$ of $P$, $$ \frac1Q:=1-\frac1P.$$ Note $P,Q\in[1,\infty]$. I now claim that $$ Q(-p)=-r.$$ Indeed, recall that $\frac1{-r} + \frac1p = 1$. So we want $1/(Q(-p))$ to equal $1-1/p$. lets check: $$\frac1{-Qp} = -\frac1p\left(1-\frac1P\right) = \frac{p-1}p = 1-1/p$$ hence, we can apply classical Holder's for the conjugate pair $P,Q$ to get $$ \int f^p \le \|(fg)^p\|_{L^{P}} \|g^{-p}\|_{L^Q} = \|fg\|_{L^1}^{1/P} \|g^{-r}\|_{L^1}^{1/Q} = \|fg\|_{L^1}^p \|g^{-r}\|_{L^1}^{p/r}$$ (We have used the property that $\|f\|^A_{L^A} = \|f^A\|_{L^1}$ twice.) Now taking $p$th roots, we obtain the claim as desired.