# Getting a lower bound for an L1 norm using higher moments

I would like to discuss a principle that came up in this recent talk of Adam Harper as well as my own research with Burak Erdogan.

Let $f : \Omega \to \mathbb{C}$ be a function on some measure space with measure $\mu$ (for instance $\Omega = [0,1]$ or a finite set).

Often one is interested in finding lower bounds for the $L^1$ norm of $f$, that is $\int_{\Omega} |f| d\mu$, but has no way to directly estimate it. As a toy example, we can consider $g_N: \mathbb{R} / \mathbb{Z} \to \mathbb{C}$ via $g_N(x) = \sum_{1 \leq n \leq N} e(x n^2)$. Estimating the $L^1$ norm directly seems hard.

But sometimes, we are able to estimate higher $L^p$ norms of $f$. This is useful to our original problem, since an application of Holder’s inequality reveals that a lower bound on the $L^2$ norm and an upper bound on the $L^4$ norm gives a lower bound on the $L^1$. To see this, note $\int |f|^2 \leq (\int |f|^4 )^{1/3} (\int |f|)^{2/3}$.

We can apply this idea to our original example. Parseval’s identity gives that $\int |g_N|^2 = N,$ while orthogonality and the divisor bound give that $\int |g_N|^4 \lesssim_{\epsilon} N^{2 + \epsilon}$. This gives $\int |f| \gtrsim_{\epsilon} N^{1/2 - \epsilon}$ and is expected by the heuristic that a typical exponential sum should be about square root of the length of the sum.

My intuition is the following. Suppose the measure space is a probability space. Then $\int |f| \leq (\int |f|^2)^{1/2}$. We are basically trying to reverse this inequality. Equality holds when $f$ is constant, that is $f$ is not too concentrated. The upper bound on the $L^4$ norm of $f$ implies that indeed $f$ is not too concentrated.

We mention that there is nothing too special about the exponents 2 and 4 chosen for the above discussion (although they are convenient for the specific example I chose).

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