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Assume $\phi: \mathbb{R}^d \to \mathbb{R}$ and has the property that $\left| \int_Q \phi dx \right| \leq M \frac{m(Q)}{(1 + m(Q))} $ holds for every rectangle $Q$ for some constant $M$ that is independent of $Q$, then $\lim_{k \to \infty} \int_{\mathbb{R}^d} \phi(kx)f(x) dx = 0$ for ever $f \in L^1(\mathbb{R}^d)$

I know that you can just verify the claim for an increasing sequence of cubes. If $Q_n = [-n,n] \times[-n,n] \times [-n,n]$ ($d$-times), then we can do a change of variables so that

\begin{equation} \int_{Q_n} \phi(kx)f(x) dx = \frac{1}{k^d} \int_{Q_{nk}} \phi(y)f(y/k) dy \end{equation}

This looks closer to what I want. If I can get some control of $f$ I could use the fact that $\frac{1}{k^d}\left| \int_Q \phi dx \right| \leq \frac{M}{k^d} \frac{m(Q)}{(1 + m(Q))} $ and as $k \to \infty$ we would get the answer. But I do not know what I can do with the function $f$.

Any thoughts or suggestions would be appreciated.

Mike
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1 Answers1

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Define the family $\mathcal{D} \subset L^1(\mathbb{R}^d)$ by

$$ \mathcal{D} = \Biggl\{ \sum_{i=1}^{n} c_i \mathbf{1}_{Q_i} : \text{$c_1, \cdots, c_n$ constants, $R_1, \cdots, R_n$ rectangles} \Biggr\}.$$

Then $\mathcal{D}$ is a dense subset of $L^1(\mathbb{R}^d)$. Moreover, by the assumption,

$$ T_k(f) = \int_{Q} \phi(kx)f(x) \, \mathrm{d}x \tag{*} $$

defines a linear functional on $\mathcal{D}$. Now let us make some observations:

  1. For each rectangle $Q$, $$ \left| T_k(\mathbf{1}_Q) \right| = k^{-d}\left| \int_{kQ} \phi(x) \, \mathrm{d}x \right| \leq \frac{M}{k^d} \frac{m(kQ)}{1+m(kQ)} \leq M m(Q). $$ In particular, the inequality in the second step tells that $T_k(\mathbf{1}_Q)$ converges to $0$ as $k \to \infty$. Then by linearity, $T_k(f) \to 0$ for any $f \in \mathcal{D}$.

  2. For $f = \sum_{i=1}^{n} c_i \mathbf{1}_{Q_i} \in \mathcal{D}$, we may assume that $Q_i$'s are non-overlapping. Then $$ \left| T_k f \right| \leq \sum_{i=1}^{n} \left| a_i \right| \left| T(\mathbf{1}_{Q_i}) \right| \leq M \sum_{i=1}^{n} \left| a_i \right| m(Q_i) = M \| f \|_{L^1}. $$ This tells that each $T_k$ extends to a continuous linear functional on $L^1(\mathbb{R}^d)$, which we also write as $T_k$. Then the extension $T_k$ is again given by $\text{(*)}$ and the inequality $\left| T_k(f) \right| \leq M\|f\|_{L^1}$ continues to hold for $f \in L^1(\mathbb{R}^d)$.

Combining altogether, for any $f \in L^1(\mathbb{R}^d)$ and $g \in \mathcal{D}$,

$$ \limsup_{k\to\infty} \left| T_k(f) \right| \leq \limsup_{k\to\infty} \left| T_k(g) \right| + M \| f - g \|_{L^1} = M \| f - g \|_{L^1}. $$

So by letting $g \to f$ in $L^1$, the desired claim follows.

Sangchul Lee
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  • How does $\limsup |T_k(g)| = 0$? Is it because in your first step you could say $T_k(1_{Q}) <k^{-d}M$? How do you know that each $T_k$ extends to a continous linear functional on $L^1(\mathbb{R}^d)?$ – Mike Nov 04 '20 at 22:44
  • @Mike, In the first step, we saw that $T_k(\mathbf{1}_Q)\to0$ as $k\to\infty$. So by linearity, the same is true for any element $g$ of $\mathcal{D}$. Next, any uniformly continuous $\mathbb{R}$-valued function defined over a dense subset of a metric space $X$ uniquely extends to a continuous function on all of $X$. (The codomain $\mathbb{R}$ can be replaced by any complete metric space, but this is not relevant to our discussion. You may check this as well.) Then by limiting argument, it is routine to check that the extension is also linear. – Sangchul Lee Nov 04 '20 at 22:49
  • I see, thank you very much for your response. – Mike Nov 04 '20 at 22:53
  • So the most nontrivial part (which I did not stress out in my solution) is why the extension is also given by the integral $\text{()}$. If $\phi$ is known to be bounded in advance, then the answer is very simple, since $T_k$ is a priori* defined on all of $L^1(\mathbb{R}^d)$. Otherwise, the duality of $L^p$-space tells that $T_k$ must be given by the form $$T_k(f)=\int_{\mathbb{R}^d}\psi(x)f(x),\mathrm{d}x$$ for some $L^{\infty}$-function $\psi$, and then it is routine to check that $\psi(\cdot)=\phi(k\cdot)$. – Sangchul Lee Nov 04 '20 at 22:55