The following small result was used in a recent paper of B. Dodson (p. 8 of this preprint, from July 2). I thought I’d provide the steps, since the text merely states that it can be seen “[b]y Littlewood–Paley arguments”.

Lemma: Let \(\Omega \in C^{\infty}(\mathbb{R}^d \setminus \{0\})\) be a smooth homogeneous function of degree zero, so that (indeed, equivalently) \(\Omega \mid_{S^{d \ – \, 1}}\) is a smooth function on the sphere. Then for any dyadic number \(M\), and any \(k \geq 1\), $$(P_{\geq M} \Omega)(x) = \Omega(x) \ – \, (P_{< M} \Omega)(x) = O_k(\langle M x \rangle^{- k}).$$

(The canonical example for \(\Omega\) should be thought-of as the map \(\Omega(x) = x / |x|\). This result shows that the high-frequency contributions for \(\Omega\) are vanishingly-small, compared to the bulk of the function which is concentrated in low-frequency, high-mass parts piled up at the frequency zero (which makes sense, given the hypothesized smoothness of \(\Omega\)). Furthermore, the decay is better away from the origin than close to it, which is a reflection of the slower convergence of the Littlewood-Paley expansion at the point where the function is irregular.)

This will be reduced to the following three lower lemmas, the first two of which are obvious. We will focus on the third.

Sub-Lemma 1: Let \(f \in L^{\infty}(\mathbb{R}^d)\). Then the ordered infinite sum \(\sum_{N \geq M} P_N f\) converges to \(P_{\geq M} f\), in the weak\(^*\)-topology on \(L^{\infty}(\mathbb{R}^d)\). (Also, as in the preprint, the operator \(P_{\geq M}\) on \(L^{\infty}\) functions is defined by duality, taking the adjoint of \(P_{\geq M} = I – P_{< M}\), which is \(L^1 \to L^1\) bounded (and self-adjoint).)

Sub-Lemma 2: Let \(\Omega \in C^{\infty}(\mathbb{R}^d \setminus \{0\})\) be a smooth homogeneous function of degree zero. Then for all \(\alpha \in \mathbb{N}_0^d\), $$|(\partial^{\alpha} \Omega)(x)| \lesssim |x|^{- |\alpha|}.$$

(For a proof, write \(\Omega(x) = \Omega(x / |x|)\) and differentiate using the chain rule, product rule, and induction.)

Sub-Lemma 3: For any \(N \in 2^{\mathbb{Z}}\), and any \(k \geq 1\), for \(\Omega\) satisfying the previous hypotheses, $$|(P_N \Omega)(x)| \lesssim_k \langle N x \rangle^{- k}.$$

By a dyadic sum and Lemma 1, it is obvious that the pointwise behavior in Lemma 3 implies the Proposition (doing this for the far-range \(x\) (those with \(|M x| \gg 1\)); for those near the origin, we simply revert to the boundedness of \(\Omega\)).

Proof: Fixing \(k\) and \(N\), we may restrict consideration to \(x \in \mathbb{R}^d\) with \(|N x| \gg 1\), and then take \(K \sim |N x|\) for the relevant large dyadic number \(K \in 2^{\mathbb{N}}\). Then we write $$(P_N \Omega)(x) = \int_{\mathbb{R}^d} N^d \check{\psi}(N y) \Omega(x \ – \, y) \, dy,$$ which we may rewrite as $$\int_{\mathbb{R}^d} \check{\psi}(y) \Omega(x \ – \, y / N) \, dy,$$ and by the degree-zero homogeneity of \(\Omega\), this is $$\int_{\mathbb{R}^d} \check{\psi}(y) \Omega(N x \ – \, y) \, dy.$$

We now use the fact that \(\check{\psi}\) has vanishing moments to all orders (a result of its Fourier transform \(\psi\) vanishing in a neighborhood of the frequency origin). We subtract off $$\int_{\mathbb{R}^d} \check{\psi}(y) \bigg( \Omega(N x \ – \, y) \ – \, \sum_{|\alpha| < k} \frac{1}{\alpha!} (\partial^{\alpha} \Omega)(N x) (- y)^{\alpha} \bigg) \, dy.$$

We now use the Schwartz nature of \(\check{\psi}\) and take a dyadic expansion by \(|y|\); this gives $$\Bigg( \sum_{J \ll K} + \sum_{J \sim K} + \sum_{J \gg K} \Bigg) \int_{|y| \sim J} ( \cdots ) \, dy$$ (with the obvious modification at \(J = 1\)). Then, applying the standard manipulations for each piece in the \(\sum_{J \ll K}\) sum, the mean-value form of Taylor’s theorem gives a bound on the magnitude of $$\lesssim_k \sum_{J \ll K} \int_{|y| \sim J} J^{- (k + 2 d)} \sup_{\theta \in [0, 1]} |(\nabla^k \Omega)(N x \ – \, \theta y)| \, dy,$$ from which we get (using Lemma 2) $$\lesssim_{\Omega} \sum_{1 \leq J \leq K / 8} J^{- k} |N x|^{- k} \lesssim K^{- k},$$ and we recognize this to be admissible.

For the terms with \(J \gtrsim K\), we now collapse the Taylor series entirely and treat it solely as a polynomial with coefficients to be controlled, plus we collapse the dyadic sum into a single integral, to get $$\lesssim \int_{|y| \gtrsim K} |\check{\psi}(x)| \bigg(1 + \sum_{|\alpha| < k} |N x|^{- |\alpha|} |y|^{|\alpha|} \bigg) \, dy$$ by Lemma 2. This is then $$\lesssim \int_{|y| \gtrsim K} \langle y \rangle^{- (2 k + d)} \langle y \rangle^{k \, – \, 1} \, dy \lesssim K^{- k},$$ and we obtain the conclusion.

---

Also, Fedor Nazarov once wrote a response on MSE, proving Hölder in Lorentz spaces using the dyadic decomposition by height (as opposed to width, the conventional way to do it). In it, he relied the following inequality: let \((a_n)_{n \in \mathbb{Z}} \subseteq (0, \infty)\) with only finitely-many nonzero terms, and take any \(\beta > 0\), \(\epsilon > 0\), and \(c, d \in \mathbb{R}\). Then $$\sum_n 2^{- \epsilon |n \ – \, c \log(a_n) \ – \, d|} a_n^{\beta} \lesssim \bigg( \sum_n a_n \bigg)^{\beta} \lesssim \sum_n 2^{\epsilon |n \ – \, c \log(a_n) \ – \, d|} a_n^{\beta}.$$

Unfortunately, he did not give any indication of the proof of this “obvious” fact. I improvised an argument, which works (after a while), but things get unbelievably messy for this simple-looking problem.

We begin with the right-hand side. By the subadditivity of fractional powers, it suffices to prove this for \(\beta > 1\). Write $$k_n = n \ – \, c \log(a_n) \ – \, d, \qquad 2^{(n \ – \, k_n \ – \, d)/c} = a_n;$$ we wish to show, for \(\beta > 1\), $$\bigg( \sum_n 2^{(n \ – \, k_n)/c} \bigg)^{\beta} \lesssim \sum_n 2^{\epsilon |k_n|} 2^{\beta (n \ – \, k_n)/c}$$ We claim it suffices to show this for \(c > 0\). Indeed, if the above holds for all sequences and all \(c > 0\), then take \(– \tilde{k}_{- n} = k_n\), then reindex the sum.

We also claim that it suffices to show this for \(k_n\) exactly integer on its set of nonvanishing. Indeed, we note that in the above expression, everything (\(2^{(n \ – \, k_n) / c}\), \(2^{\epsilon |k_n|}\), \(2^{\beta (n \ – \, k_n) / c}\)) shifts only by a multiplicative constant which is absolute (depending only on \(\beta\), \(c\)) and two-sided, if the value of \(k_n\) is shifted by something \(O(1)\).

First we bound the contribution of the indices with \(|k_n| \leq |n| / 2\). In this case, denoting the relevant set by \(A\), we have $$\sum_{n \in A} 2^{(n \ – \, k_n) / c} = \sum_j \sum_{\substack{n \in A \\ \Phi(n) = j}} 2^{\Phi(n) / c} = \sum_j|\{\Phi \mid_A = j\}| 2^{j / c}.$$

Now, we note that if \(j = n \ – \, k_n = n’ \ – \, k_{n’}\), then this means \(n \ – \, j = k_n\) for all \(n \in \Phi^{- 1}(\{j\})\). As a consequence, we get the right-hand side to be $$\sum_{j \in \Phi(A)} \sum_{\Phi(n) = j} 2^{\epsilon |k_n|} 2^{\beta (n \ – \, k_n) / c} = \sum_{j \in \Phi(A)} \sum_{\Phi(n) = j} 2^{\epsilon |n \ – \, j|} 2^{\beta j / c}.$$

By recognizing and summing the geometric series in \(n\), this is $$\simeq_{\epsilon} \sum_{j \in \Phi(A)} 2^{\epsilon |k_{n_j}|} 2^{\beta j / c},$$ where \(n_j \in \{\Phi = j\}\) maximizes \(n \mapsto |n \ – \, j| = |k_n|\) on the set \(\{\Phi = j\}\). In other words, $$|k_{n_j}| = \max_{n \in \{\Phi = j\}} |k_n|$$ for \(j \in \Phi(A)\).

We want to show that $$\bigg( \sum_{j \in \Phi(A)} |\{\Phi \mid_A = j\}| 2^{j / c} \bigg)^{\beta} \lesssim \sum_{j \in \Phi(A)} 2^{\epsilon |k_{n_j}|} 2^{\beta j / c}.$$

To complete this, we note that any level set of size \(s_j = |\{\Phi \mid_A = j\}| > 1\) in \(A\) will have an element of the set with \(|n \ – \, j| \gtrsim s_j / 2\), since the set has diameter \(\gtrsim s_j\).

We now write the \(\ell^1\) sum as $$\sum_{r \geq 1} \sum_{\substack{j \in \Phi(A) \\ s_j = r}} r 2^{j / c} \lesssim \sum_{r \geq 1} r 2^{\max(E_r) / c},$$ where $$E_r = \{j \in \Phi(A) : |\{\Phi \mid_A = j\}| = r\} = \{j \in \Phi(A) : s_j = r\}.$$

Thus we have that the \(\beta\)th power of the sum as $$\lesssim \sum_{r \geq 1} 2^{\beta \max(E_r) / c} 2^{\epsilon |k_{n_{\max(E_r)}}|} \bigg( \sum_{r \geq 1} r^{\beta’} 2^{- \epsilon (\beta’ / \beta) |k_{n_{\max(E_r)}}|} \bigg)^{\beta / \beta’}.$$

By our above remarks, we saw that when \(j\) was associated to \(s_j = |\{\Phi \mid_A = j\}| > 1\), then \(|k_{n_j}| \gtrsim s_j\), and so for \(j = \max(E_r)\), we get $$|k_{n_{\max(E_r)}}| \gtrsim s_{\max(E_r)} = r.$$ This means we have the internal factor of$$\lesssim \bigg( \sum_{r \geq 1} r^{\beta’} 2^{- \eta r} \bigg)^{\beta / \beta’} \lesssim 1.$$

This finally completes the damn estimate for \(n \in A\). For \(n \in B = \{j \in \mathbb{Z} : |k_j| > |j| / 2\}\), things are much easier, since we can simply write the \(\beta\)th power of the \(\ell^1\) sum as $$\leq \sum_{n \in B} 2^{\epsilon |k_n|} 2^{\beta (n \ – \, k_n) / c} \bigg( \sum_{n \in B} 2^{- \epsilon (\beta’ / \beta) |n| / 2} \bigg)^{\beta / \beta’}$$ with the obvious uniform bound on the internal coefficient.

Now, we do the left-hand inequality. Here we have the goal of $$\sum_n 2^{- \epsilon |k_n|} 2^{\beta (n \ – \, k_n)/c} \lesssim \bigg( \sum_n 2^{(n \ – \, k_n)/c} \bigg)^{\beta}.$$ Again, we may assume \(c > 0\).

Here, if \(\beta \geq 1\), we can simply use the nestedness of the \(\ell^p\) spaces to conclude. So the important case is \(0 < \beta < 1\).

Take \(B = \{n \in \mathbb{Z} : |k_n| > |n| / 2\}\), and write \(\beta = 1 / p\) for a \(1 < p < \infty\). Then we can write the \(1 / \beta\)th power of the left-hand term as $$\leq \bigg( \sum_{n \in B} 2^{- \epsilon |n| / 2} 2^{(n \ – \, k_n) / (c p)} \bigg)^p \leq \sum_{n \in B} 2^{(n \ – \, k_n) / c} \bigg( \sum_{n \in B} 2^{- \epsilon p’ |n| / 2} \bigg)^{p / p’}.$$

For the remaining part, take \(A = \{n \in \mathbb{Z} : |k_n| \leq |n| / 2\}\). The problem is equivalent to $$\bigg( \sum_{n \in A} 2^{(n \ – \, k_n) / (c p)} 2^{- \epsilon |k_n|} \bigg)^p \lesssim \sum_{n \in A} 2^{(n \ – \, k_n) / c},$$ so we write the right-hand side as $$\sum_{n \in A} 2^{(n \ – \, k_n) / c} = \sum_r \sum_{\substack{n \in A \\ k_n = r}} 2^{(n \ – \, k_n) / c} \simeq \sum_r 2^{(n_r \ – \, r) / c},$$ where obviously \(n_r = \max(n \in A : k_n = r)\).

Likewise, the left-hand sum can be written as $$\sum_r \sum_{\substack{n \in A \\ k_n = r}} 2^{(n \ – \, r) / (c p)} 2^{- \epsilon |r|} \simeq \sum_r 2^{(n_r \ – \, r) / (c p)} 2^{- \epsilon |r|}.$$

Hence we wish to get $$\bigg( \sum_r 2^{(n_r \ – \, r) / (c p)} 2^{- \epsilon |r|} \bigg)^p \lesssim \sum_r 2^{(n_r \ – \, r) / c},$$ but this inequality is (mercifully) immediate.

From this we conclude the two-sided estimate.

---

The following argument just came to me this Tuesday, after realizing a question I’d been wondering about for a while is trivial:

We know that \(L^{1, 1} = L^1\) is always normable, and \(L^{1, \infty}\) is in-general not normable. What about \(L^{p, q}\) in-between? Those with \(p = 1\), \(1 < q < \infty\)?

These are not normable either. Indeed, suppose otherwise; then \((L^{1, q})’\), as a normed linear space, would have a dual. In particular, there exists a norming set of linear functionals on \(L^{1, q}\).

By the nestedness of Lorentz spaces, we have \(L^{1, 1} = L^1 \hookrightarrow L^{1, q}\). It follows that linear functionals \(\ell\) would obey \(|\ell(f)| \leq \|f\|_{L^{1, q}} \lesssim \|f\|_{L^1}\).

Now, assuming \((X, \mathcal{M}, \mu)\) satisfies the standard \(\sigma\)-finiteness assumption, we have that \(\ell = \ell_g\) is represented by an element \(g \in L^{\infty}\) on \(L^1 = L^{1, 1} \cap L^{1, q} \hookrightarrow L^{1, q}\).

Now, multipliers with characteristic functions form contractions of the \(L^{1, p}\) norm, and so (by restricting to the sets \(\{\Re( \)latex \ell_{g_{+}}$, \(\ell_{g_{-}}\), and thus \(\ell_{|g|}\) are bounded and belong to \((L^{1, q})’\).

Because \((L^{1, q})’\) has a norming collection, we may assume without loss that \(\ell\) is nontrivial, and moreover, by truncating elements \(f \in L^{1, q}\) to \(f \chi_{\{N^{- 1} < |f| < N\}}\) for some large \(N \in 2^{\mathbb{Z}}\), we get that \(f \ – \, f_N \to 0\) in the standard (measure-theoretic) \(\| \cdot \|_{L^{1, q}}^*\) quasinorm (using, say, dominated convergence). Hence we have density of \(L^1\) in \(L^{1, q}\), and so we get that \(g\) is nontrivial.

We now specify \((X, \mathcal{M}, \mu) = (\mathbb{R}, \mathcal{B}(\mathbb{R}), \lambda^1)\). Since \(g\) is nontrivial, it has Lebesgue points, and we now construct a function adapted to such a point \(x_0\) as follows (where, without loss, \(|g(x_0)| = 1\) and \(x_0 = 0\)): we have $$\int_{r < |t| < R} |g(t)| \, dt = 2 (R \ – \, r) \ – \, o(r + R)$$ for all \(k\), which enables us to define \(f\) set to equal \(\frac{1}{k} 2^k\) on the ring \(|t| \sim 2^{- k}\). Then we see that we can make \(o(R_k) = o(2 \cdot 2^{- k})\) can be made much smaller than the length of the interval.

Now using the Lorentz norm by width (height could be done, but the presence of the \(\frac{1}{k}\) corrections on the \(2^k\) means the computation becomes a complicated mess presenting the transcendental inverse of the function \(k \ – \, \log(k)\)), we have $$\|f\|_{L^{1, q}}^q \lesssim \|(2^k / k \cdot 2^{- k / 1})_k\|_{\ell^q(\mathbb{N})}^q = \sum_{k \geq 1} k^{- q} < \infty,$$ and yet $$|\langle f_N, g \rangle| = \sum_{k = 1}^{N} \frac{1}{k} 2^k \cdot \frac{1}{2} 2^{- k} \geq \log(N) \ – \, O(1).$$ This last quantity, of course, is $$|\ell_{|g|}(f_N)| \leq 2 \|f_N\|_{L^{1, q}} \lesssim 2 \|f_N\|_{L^{1, q}}^* \leq 2 \|f\|_{L^{1, q}} \lesssim_q 1.$$

This contradiction shows that no equivalent norm on \(L^{1, q}\) can exist in general.