Let $ X_N = \text{span} \{\cos(2\pi lx): l=0, \cdots, N-1 \} $ with $ x \in [0, 1] $ and $ Y_N = \{v =(v_0, \cdots, v_{N-1}): v_j \in \mathbb{C}\} = \mathbb{C}^N $. Then $ X_N $ is the space of trigonometric polynomials. We equip $ X_N $ with the usual $ L^p (1 \leq p \leq \infty ) $ norm and equip $ Y_N $ with the discrete $ l^p (1 \leq p \leq \infty)$:

$$ \| v \|_{l^p} = \left( \frac{1}{N}\sum_{j=0}^{N-1} |v_j|^p \right)^\frac{1}{p}, \quad 1 \leq p < \infty, $$

$$ \| v \|_{l^\infty} = \max_{0 \leq j \leq N-1} |v_j|, \quad v \in Y_N. $$

Let $ R_N:X_N \rightarrow Y_N $ be the restriction of the trigonometric polynomials to some discrete points: for $ u \in X_N $,

$$ v = R_N u, \quad v_j = u(x_{j+1/2}), \quad j=0, \cdots, N-1, $$ where $ x_{j+1/2} = (j+1/2)/N $.

Question: can we find a constant $ C $ independent of $ N $ such that

$$ \| R_N u \|_{l^4} \leq C\| u \|_{L^4}, \quad u \in X_N. $$

Moreover, does it hold for higher dimensions though currently it is only stated in 1D?

Note that by Parseval's identity, one has,

$$ \| R_N u \|_{l^2} = \| u \|_{L^2}. $$


1 Answer 1


Yes, this goes back to the work of

Plancherel, M.; Pólya, George, Fonctieres entières et intégrales de Fourier multiples, Comment. Math. Helv. 9, 224-248 (1937). ZBL0016.36004.

(see for instance Theoreme III). Nowadays one would usually prove such a result using Littlewood-Paley projections or similar devices to express $R_N u$ as a suitable integral expression of $u$ to which Schur's test (for instance) can be applied.

  • $\begingroup$ Thanks. I'm also confused about how the validity of the result depends on the discrete points selected. Is the set of points that guarantees the equality of l^2 norm and L^2 norm the only possible choice or is it not that essential? $\endgroup$
    – Ghoshee
    Nov 3, 2022 at 5:33
  • $\begingroup$ As long as the points sampled are separated from each other by at least a constant multiple of the natural spatial uncertainty (in this case, $1/N$), one obtains an upper bound of this form (one can think of this as a manifestation of the uncertainty principle). The lower bound is more interesting (one needs the sampling points to be at least as dense as the Shannon sampling threshold) and becomes quite delicate if the sampling points are not equally spaced (deep theorems such as Beurling-Malliavin become relevant). $\endgroup$
    – Terry Tao
    Nov 3, 2022 at 16:56

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