Can we find a function satisfiying the following functional inequation?

Question:

I would be interested know if there exist (and construct)

• a function $f:\mathbb N\times \mathbb N \to (0,\infty )$ such that for all integers $p,q$, $f(p,q)=f(q,p)$ and

• a constant $C>0$,

such that, for all $p,p',q,q',k\in\mathbb N$ with $k\leq p$ and $k\leq q'$, $$\frac{p!q'!}{k!(p-k)!(q'-k)!}\frac{f(p+p'-k,q+q'-k)}{f(p,q)f(p',q')}\leq C\,.$$ Do you have any idea?

EDIT:

This problem has no solution, see the comments.

Remarks:

• The assumption $f(p,q)=f(q,p)$ could eventually be dropped.
• Until now I just tried a naive approach by trying some functions as $f(p,q)=p!q!$, $f(p,q)=2^{p+q}$, $f(p,q)=2^{-p-q}$, etc., and got nothing very interesting.

Motivation:

This problem comes from the consideration of the algebra $$\mathcal A:=\bigoplus_{p,q\geq0}\mathcal L (\mathcal Z^{\otimes_sp},\mathcal Z^{\otimes_sq})$$ for a product $\sharp$ called the Moyal product. I was interested in building a norm for $\mathcal A$ which would make it a Banach algebra. And where the norm would have been of the form $$\|(x_{p,q})_{p,q}\|_\mathcal A := \sum_{p,q} f(p,q)\|x_{p,q}\|_{\mathcal L (\mathcal Z^{\otimes_sp},\mathcal Z^{\otimes_sq})}$$ for $(x_{p,q})_{p,q}\in\mathcal A$.

For a Banach algebra you have the condition $$\|(x_{p,q})_{p,q}\sharp (x'_{p,q})_{p,q}\|\leq \|(x_{p,q})_{p,q}\| \,\|(x'_{p,q})_{p,q}\|\,$$ but in fact it is enough to construct a norm such that $$\|(x_{p,q})_{p,q}\sharp (x'_{p,q})_{p,q}\|\leq C\|(x_{p,q})_{p,q}\| \,\|(x'_{p,q})_{p,q}\|\,$$ for some constant $C>0$ to know that a norm satisfying the condition above exists.

And that is this last relation expressed in terms of the coefficients $f(p,q)$ which led to the problem above.

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Could you add a little more detail or motivation for why you're interested in such a function? That might help point towards a class of functions which satisfy the inequality. –  Gwyn W Nov 28 '12 at 22:00
@GwynW Well in fact I solved the problem myself, there is no solution. To see it consider the case $p=q=p'=q'$ and set $g(p)=f(p,p)$. Choosing $k=p$ gives $$(a) : p!\leq Cg(p).$$ Choosing $k=0$ gives $g(2p)\leq C g(p)^2$. From that we deduce $$(b) : g(2^n)\leq C^{2^n-1}g(0)^{2^n}.$$ And one then gets a contradiction from (a) and (b). But thanks for your help I will give some motivation. –  Sebastien B Nov 28 '12 at 22:12