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.)

  • $\begingroup$ $[0,1]$ 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 $\|p\|_1 \leq N^2/\|p\|_4$. $\endgroup$ Oct 27, 2017 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. $\endgroup$ Oct 27, 2017 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? $\endgroup$ Oct 27, 2017 at 13:10

2 Answers 2


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 $[N/2, 2N]$.

  • $\begingroup$ 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? $\endgroup$ Oct 27, 2017 at 16:57
  • 1
    $\begingroup$ 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, 2017 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 $\sqrt{N}$. $\endgroup$ Oct 27, 2017 at 17:16

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.