Let $p(x) = \sum\_{n=1}^N e^{2 \pi i a\_n x}$ be a trigonometric polynomial, where $a\_n$ are distinct positive integers. There is a classical trick which (using Hölder's inequality) allows to give a lower bound for the $L^1$ norm of $p$ in terms of the $L^4$ norm of $p$. One obtains
$$
\|p\|\_1 \geq \frac{\|p\|\_2^{3}}{\|p\|\_4^{2}} = \frac{N^{3/2}}{\|p\|\_4^{2}}.
$$
So roughly speaking a small $L^4$ norm implies a large $L^1$ norm.


**Question:** Is the opposite also true? That is, does a large $L^4$-norm imply a small $L^1$-norm? (And if "yes", is there a quantitative estimate?)


(This might be a stupid question, but still I am grateful for an answer.)


(PS: For a reference to the trick mentioned above, see for example A. A. Karatsuba, "An estimate of the L1-norm of an exponential sum", Mathematical Notes, 1998, 64:3, 401-404.)



$L^1$ and $L^4$ norms of trigonometric polynomials
No this need not be the case. Take $f(x) = \sum\_{n=1}^{N/2} e(nx)$ and $g(x) = \sum\_{k=N/2}^N e(2^kx)$. Then the $L^4$ norm of $f$ is big -- of size $N^{\frac 34}$ -- and its $L^1$ norm is very small -- of size $\log N$. On the other hand the $L^4$ norm of $g$ is small -- of size $\sqrt{N}$ -- and its $L^1$ norm is correspondingly large -- of size $\sqrt{N}$. But now the triangle inequality shows that the $L^4$ norm of $f+g$ is big (of size $N^{\frac34}$), whereas the $L^1$ norm of $f+g$ is also big (of size $N^{\frac 12}$). 


Even if you want the coefficients $a\_n$ to be small, you could arrange this by making the first half of the coefficients be all natural numbers in $[1, N/2]$ and then choosing $N/2$ integers randomly from $[N/2, 2N]$. 


