3

Besicovitch space is a space constructed in the following way:

  1. We take the closure (with respect to the uniform convergence topology) of a linear span: $B_0=\overline{\operatorname{span}\{\lambda\in\mathbb R: e^{i\lambda x}\}}$
  2. We define an inner product as $(f,g)=\lim\limits_{T\to\infty} \frac{1}{2T}\int\limits_{-T}^T f(x)\overline{g(x)}dx$
  3. We take quotient space $B_i=B_0\setminus\ker\sqrt{(f,f)}$
  4. We take space $B_2$ as a completion of $B_1$ by a norm induced by the inner product

I have some questions:

If the limit in 2) exists, all axioms of an inner product are satisfied. But how can I show that this limit exists?

Davide Giraudo
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Тимофей Ломоносов
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1 Answers1

1

First consider the case where $f(x)=e^{i\lambda x}$ and $g(x)=e^{i\lambda' x}$. In this case, $f(x)g(x)=\exp(i(\lambda-\lambda')x)$ and we can compute explicitely the integral.

The result holds by linearity when $f$ and $g$ are linear combinations of $e^{i\lambda \cdot}$. Then we use an approximation argument: let $f_N$, $g_N$ of this form such that $\lVert f-f_N\rVert_\infty+\lVert g-g_N\rVert_\infty\lt 1/N$. Then $$\left|\frac 1{2T}\int_{-T}^Tf(x)\overline{g(x)}dx-\frac 1{2S}\int_{-S}^Sf(x)\overline{g(x)}dx\right|\leqslant C/N+\left|\frac 1{2T}\int_{-T}^Tf_N(x)\overline{g_N(x)}dx-\frac 1{2S}\int_{-S}^Sf_N(x)\overline{g_N(x)}dx\right|,$$ where $C$ is independent of $N$, $S$ and $T$. We thus have $$\limsup_{S,T\to \infty}\left|\frac 1{2T}\int_{-T}^Tf(x)\overline{g(x)}dx-\frac 1{2S}\int_{-S}^Sf(x)\overline{g(x)}dx\right|\leqslant C/N.$$

Davide Giraudo
  • 181,608