$\left| \int_Q \phi dx \right| \leq M \frac{m(Q)}{(1 + m(Q))} $ for every cube, then $\lim_{k \to \infty} \int \phi(kx)f(x) dx = 0$ for $f \in L^1$ 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.
 A: 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:

*

*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 $f \in \mathcal{D}$.


*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.
