Functions vanishing to infinite order "in $1$-mean" When proving certain unique continuation theorems for classes of functions which satisfy some type of PDE inequality, people often talk about functions which vanish to infinite order at a point $x_0$ in the $q$-mean in the sense that, if $u: \Omega \to \mathbb{C^n}$ is an $L^q$-function, where $\Omega \subset \mathbb{R}^n$ is an open set: $$\int_{|x - x_0| \leq r} |u(x)|^q dx = O(r^k), \hspace{5pt} \forall k \in \mathbb{N}.$$By Hölder's inequality, we know that if $u$ vanishes to infinite order in the $q$-mean, and $p<q$, then $u$ also vanishes to infinite order in the $p$-mean.
Vanishing to infinite order in the sense that we're used to for smooth functions (i.e, $|u(x)| = O(|x-x_0|^k)$ when $ x \to x_0$ for all $k$; this makes sense even for non-smooth functions) is equivalent to vanishing to infinite order in the $p$-mean for $p = \infty.$
My question can be phrased in two ways:

Does the converse hold? I.e, if $p<q$, and $u$ vanishes to infinite order in the $p$-mean, does it also vanish to infinite order in the $q$-mean?


A more specific version of the question which I'm interested in, for the purpose of $J$-holomorphic curves: if $u$ is a smooth function and $n=2$ (for example, $\Omega$ can be a small open disk around $0$), does vanishing to infinite order in the $1$-mean imply vanishing to infinite order in the usual sense, i.e. $D^{\alpha}u(0) = 0$ for all multi-indices $\alpha$? This is stated in Section 2.3. of the McDuff-Salamon book "J-holomorphic curves and symplectic topology." I was able to show this for $|\alpha| = 0,1$, but I wasn't able to solve the general case, where I get stuck at an expression of the form $$\int_{|z| \leq r} |(D^k u)_0(z,\dots,z)|dz \leq M_Nr^N, \hspace{7pt} \forall N, \forall r<r_0.$$I was also able to show that it's true when $u$ is holomorphic, due to the specific nature of the $k$th derivative in this case.

 A: For simplicity, I work in two dimensions.
Smooth case : If $u$ is assumed to be smooth in a neighborhood of $0$, then indeed, its vanishing to infinite order in 1-mean (and a fortiori, in $q$-mean for $q > 1$) implies that all derivatives of $u$ at $0$ vanish. This follows from the Taylor expansion with remainder, which states that for all $N \ge 0$,
$$ u(x,y) = \sum_{0 \le k+l \le N} \frac{1}{k! \, l!} \frac{\partial^{k+l} u}{\partial^k x \partial^l y}(0,0) \; x^k y^l + O((x^2 + y^2)^{\frac{N+1}{2}}) \, . $$
Let's assume by induction that for $N \ge 0$, it is known that all derivatives of $u$ at $(0,0)$ up to order $N-1$ vanish, so that
$$ u(x,y) = \sum_{k+l = N} \frac{1}{k! \, l!} \frac{\partial^{k+l} u}{\partial^k x \partial^l y}(0,0) \; x^k y^l + O((x^2 + y^2)^{\frac{N+1}{2}}) \, =: u_N(x,y) + O((x^2 + y^2)^{\frac{N+1}{2}}). $$
We aim to prove that $u_N = 0$. For $r > 0$, set $\epsilon = r/\sqrt{2}$ and let $\lambda \in (0,1]$. Note that the rectangle $R(\epsilon, \lambda) := [0, \epsilon] \times [0, \lambda \epsilon]$ lies inside the ball $B(r)$ of radius $r$ centred at $0$. Integrating $u_N$ over this rectangle yields
$$ \iint_{R(\epsilon, \lambda)} u_N(x,y) dx dy = \epsilon^{N+2} \, \sum_{l=0}^{N} \frac{1}{(N-l+1)! \, (l+1)!} \frac{\partial^{N} u}{\partial^{N-l} x \partial^l y}(0,0) \; \lambda^{l+1} = P_N(\lambda) \, r^{N+2}  $$
where $P_N(\lambda)$ is a polynomial which vanishes identically in $\lambda$ if and only if $u_N = 0$. Next, taking into account that $u$ vanishes to order $N+3$ in 1-mean, we estimate that
$$ \begin{align} \left| P_N(\lambda)  \right| r^{N+2} &= \left| \iint_{R(\epsilon, \lambda)} u_N(x,y) dx dy \right| \\
& \le \left| \iint_{R(\epsilon, \lambda)} u(x,y) dx dy \right| + \left| \iint_{R(\epsilon, \lambda)} O((x^2 + y^2)^{\frac{N+1}{2}}) dx dy \right| \\
& \le \iint_{B(r)} \left| u(x,y) \right| dx dy  +  \iint_{B(r)} \left| O(s^{N+1}) \right| s ds d\theta \\
& \le O(r^{N+3}) + O(r^{N+3}) = O(r^{N+3}) \, .
\end{align} $$
Hence $|P_N(\lambda)|$ is $O(r)$ and also independent from $r$, so it vanishes identically in $\lambda$. This implies $u_N = 0$.
Continuous case: Assuming instead $u$ to be merely continuous, $u$ could vanish to infinite order at the origin in 1-mean but not in "$\infty$-mean". Let $\chi : \mathbb{R}^2 \to [0,1]$ be a continuous function supported in $B(1)$ and satisfying $\chi(0,0) = 1$. Set $$u(x,y) = \sum_{n=2}^{\infty} \frac{1}{n} \, \chi (2^n(x-1/n), 2^ny)$$
(This is a sum of functions supported in (disjoint) balls $B_n$ where $B_n$ is centred at $(1/n, 0)$ and has radius $1/2^n$.) It follows that $\mathrm{max}_{B(r)} |u| \sim r$, hence $u$ does not vanish to infinite order at the origin in the "$\infty$-mean". However, it $L^1$-norm over a ball of radius $1/N$ is of order $$\sum_{n \ge N} \frac{n}{4^n} \le  \sum_{n \ge N} \frac{1}{2^n} = \frac{1}{2^{N-1}} = O((1/N)^K) \qquad \forall K > 0.$$
However, to vanish to infinite order in the 1-mean implies vanishing to infinite order in all other $q$-mean with $q \in (1, \infty)$. First, continuity implies $u(0,0) = 0$. Indeed, assuming instead $|u(0,0)| = L > 0$, continuity of $u$ would imply that there is $r_0 > 0$ such that $|u| > L/2$ on $B(r_0)$ and hence $\int_{B(r)}|u| dxdy > (\pi L/2)r^2$ for $r < r_0$, in contradiction with $u$ vanishing at infinite order. Consequently, there exists $r_1 > 0$ such that $|u| < 1$ on $B(r_1)$, whence $\int_{B(r)}|u|^q dx dy \le \int_{B(r)} |u| dxdy = O(r^K)$ for all $K > 0$ and all $r < r_1$. Thus $(\int_{B(r)} |u|^q dxdy)^{1/q} = O(r^{K/q})$ for all $K > 0$.
General case: Without any specific constraint on $u$ besides being $L^p$ for every $p \in [1, \infty]$, I don't expect its vanishing to infinite order at $(0,0)$ in 1-mean to imply its vanishing to infinite order in $q$-mean for all other $q \in (1, \infty)$. Indeed, as a clue for this, let's consider the same function $\chi$ as above and set
$$u(x,y) = \sum_{n=2}^{\infty} 2^n \, \chi (2^n(x-1/n), 2^ny).$$
Then $u$ again vanishes to infinite order in 1-mean, but it does not even have a well-defined 2-mean. So there might be a more clever construction of a function $u$ whose $q$-means exist for all $q$, and which vanishes to infinite order in $1$-mean but not in every other $q$-means.
