Gowers norm - gap between $U_3$ and $U_4$ norms? For a function $f$, it is known that $|f|_{U_2} \le |f|_{U_3} \le |f|_{U_4} \le\dots$
Is there an example for a function $f$ such that $|f|_{U_3} < |f|_{U_4}$ (i.e. they are not equal?). The bigger the gap between them, the better.
I think there is a known construction of Gowers doing it but have no reference.
 A: Actually, I think any non-constant function would provide an example of strict inequality: the equalities you mention are proven by repeated application of the Cauchy-Schwarz inequality applied to a mixture of the function $f$ with the constant function 1, and equality in the conventional CS inequality would only hold if $f$ was also a constant: hence if not at some point in your proof the inequalities would become strict.
For the best possible gap, you should probably look at (in your case) cubic phase functions such as $e(x^3)$ - explicit calculation shows that this has large $U^4$ norm, but since it doesn't correlate well with any quadratic phase function it can't have  a large $U^3$ norm (by the inverse $U^3$ result due to Green and Tao).  The details of the calculation and how large the gap is would depend on which group you were taking the norms over.
EDIT: To be more precise, and to give a reference, the following is Exercise 11.1.12 in Tao and Vu:
If $P$ is a polynomial of degree $d$ with coefficients in $\mathbb{F}_p$ and $f(x)=e(P(x)/p)$ (defining $x\mapsto x/p$ so that it maps into $[0,1)$ in the obvious way) then we have
$$\| f\|_{U^{d'}}=1$$
for all $d'>d$ but
$$ \| f\|_{U^{d'}}\leq ((d-1)/p)^{1/2^d}$$
for all $1\leq d'\leq d$. 
For instance, if $f(x)=e(x^3/p)$ then $ \| f\|_{U^{4}}=1$ but $ \| f\|_{U^{3}}\leq (2/p)^{1/8}$ which can be made arbitrarily small by increasing $p$. 
