The integral inequality

Question

Let $f$ be an entire function of exponential type. Does the inequality $|f(a)| \le C \int_{a-1/2}^{a+1/2}|f(x)|\,dx$ hold for every $a \in R$ with an absolute constant $C$? At most, the constant may depend on $f$. The question arises in connection to spectral theory.

Answer 1

No, such an inequality need not hold: one can construct $f$ of exponential type and a sequence $\{a_n\}$ of real numbers such that $$ \frac1{f(a_n)} \int_{a_n - \frac12}^{a_n - \frac12} \left|\phantom.f(x)\right|\phantom. dx \rightarrow 0. $$ Indeed if $\{a_n\}$ increases rapidly enough then the growth of $f$ can be arbitrarily slow given that $f$ cannot be a polynomial; for example, taking $a_n = 10^n$ in the construction below makes $$ f(z) \ll \exp\left(B \phantom. \log^2 (1+\left|z\right|)\right) $$ for some absolute constant $B$ (and all $z \in {\bf C}$).

(The following construction spells out what's in Fedja's and my comments, but neither of us got around to writing it up two months ago, and now mathoverflow brought it back to the front of the queue, presumably for lack of an upvoted or accepted answer.)

The idea is to make $f(a_n)$ smaller than usual given the growth of $f$, but still larger than its average on $\left|x-a_n\right| \leq \frac12$, due to $n$-th order zeros at the edge of that interval. If $a_n \rightarrow\infty$ fast enough then $f$ can still have exponential or even much slower growth.

Let $\lbrace a_n \rbrace$, then, be a rapidly growing sequence, say $a_n = 10^n$; and define $f$ as the real Weierstrass product $f = \prod_{m=1}^\infty f_m^m = f_1 \phantom. f_2^2 \phantom. f_3^3 \phantom. f_4^4 \cdots $ where $$ f_m(x) = \Bigl( 1 - \frac{x}{a_m - \frac12} \Bigr) \phantom. \Bigl( 1 - \frac{x}{a_m + \frac12} \Bigr) $$ is the quadratic polynomial with roots at $a_m \pm 1/2$ such that $f_m(0) = 1$. Even with the growing multiplicities of the roots of $f$, the zeros are sparse enough to assure convergence and slow growth of the product.

Now for large $n$, if we restrict $x$ to $\left|x-a_n\right| \leq \frac12$ then all the factors $f_m^m$ for $m\neq n$ are essentially constant on that interval, so $f(x)$ is very nearly $\phantom.f(a_n)\phantom. \left(\phantom.f_n(x)\left/f_n(a_n)\right.\right)^n$. Thus $$ \frac1{f(a_n)} \int_{a_n - \frac12}^{a_n - \frac12} \left|\phantom.f(x)\right|\phantom. dx \sim \int_{a_n - \frac12}^{a_n - \frac12} \left(\frac{f_n(x)}{f_n(a_n)}\right)^n \phantom. dx = \int_0^1 \bigl(4u(1-u)\bigr)^n \phantom. du, $$ where $u = x - (a_n - \frac12)$. The integral is $(2^n n!)^2 \left/ (2n+1)! \right. = O(n^{-1/2}) \rightarrow 0$, QED.

Answer 2

Asking for an inequality for entire functions which are NOT polynomials makes no sense. Take Fedja's polynomial counterexample (or any other polynomial counterexample) and multiply it by $\exp(ix)$. For a counterexample to the second question, take an appropriate product.