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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.)

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  • $\begingroup$ $[0,1]$ has measure $1$, so $\|p\|_1 \leq \|p\|_4 \leq \|p\|_{\infty} \leq N$. In particular, $\|p\|_1 \|p\|_4 \leq N^2$, so that $\|p\|_1 \leq N^2/\|p\|_4$. $\endgroup$ – Peter Humphries Oct 27 '17 at 13:06
  • $\begingroup$ Well, yes , this is true, but is even weaker than the trivial $\|p\|_1 \leq \|p\|_2 \leq \sqrt{N}$. I was hoping for something stronger. $\endgroup$ – Kurisuto Asutora Oct 27 '17 at 13:09
  • $\begingroup$ I wonder if you can make use of the fact that $\|p\|_4^4 = \langle p^2, p^2\rangle$ and spectrally expand? $\endgroup$ – Peter Humphries Oct 27 '17 at 13:10
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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 $[N/2, 2N]$.

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  • $\begingroup$ Yes, I agree, this is a good example and gives a negative answer to the problem. However, in this example the polynomial decomposes in a very natural way into a "small in L^1" component (which is large in L^4) and a "small in L^4" component (which is large in L^1). Is such a decomposition always possible, or only in this particular example? $\endgroup$ – Kurisuto Asutora Oct 27 '17 at 16:57
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    $\begingroup$ A set with a large $L^4$ norm is said to have large additive energy. There are results like Balog-Szemeredi-Gowers which will allow you to find a large subsets that looks like a generalized arithmetic progression. This doesn't fully answer what you asked, but might give some ideas ... $\endgroup$ – Lucia Oct 27 '17 at 17:04
  • $\begingroup$ Yes, I know about this Balog-Szemeredi-Gowers things. But as far as I know they only give a relevant result if the $L^4$ norm (that is, additive energy) is close to the maximal possible value. I was hoping for something which also applies when the $L^4$ norm is close to $\sqrt{N}$. $\endgroup$ – Kurisuto Asutora Oct 27 '17 at 17:16
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I could only provide a approach to the problem because the exactly calculate is complex.

First,let us recall what is the equivalent condition of Holder inequality: $f\in L^p(\Omega),g\in L^q(\Omega).||f||_p||g||_q\geq ||fg||_1$. this is just:$\exists \lambda\in R,\frac{|f(x)|^p}{|g(x)|^q}=\lambda,\forall x\in \Omega$.

This inequality seem to tell us if we can construct $f,g$ with $\lambda(x)=\frac{|f(x)|^p}{|g(x)|^q}$ oscillation very frequently,i.e. $V(f,g)=\int_{\Omega}|\lambda(x)-(\frac{1}{\mu(\Omega)}\int_{\Omega}|\lambda(y)|)dy|dx$ is very large.and $||f||_p||g||_q=c(f,g)||fg||_1$,then $c(f,g)$ will be very large and lead to a counterexample or the worst situation of the original problem.

In the situation we consider,$\Omega=T_1$,$\mu$ is the canonical lesbegue measure on $T_1$.$p(x)=\sum_{n=1}^Ne^{2\pi ia_nx}$. by some calculate,the inequality $||p||_1||p||_4^2\geq ||p||_2^3$ become: $(\int_{T_1}(|p(x)|^{\frac{2}{3}})^{\frac{3}{2}}dx)^{\frac{2}{3}}(\int_{T_1}(|p|^{\frac{4}{3}})^3)^{\frac{1}{3}}\geq \int_{T_1}|p|^2$. so we need to control $V(|p|^{\frac{2}{3}},|p|^{\frac{4}{3}})$.where $\lambda(x)=\frac{|p(x)|}{|p(x)|^4}$.

fortunately we can estimate the upper bound of $V(|p|^{\frac{2}{3}},|p|^{\frac{4}{3}})$(even to get the shape estimate is possible) under the setting,because $p(x)=\sum_{n=1}^Ne^{2\pi ia_nx}$ so the ingredient of $p(x)$ is well controlled,that just mean $\lambda$ could not concentrate in a small range with a large density.

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