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).

1 thought on “Getting a lower bound for an L1 norm using higher moments

  1. Pingback: George Shakan

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