$\deg A>1$ and we prove the assertion. The following argument is the same as the proof of [@Schwede2 Theorem 3.3].

Fix a point $Q\in X$. By Lemma \[trace-diagram\], we obtain the following commutative diagram:

[$$\begin{CD}
H^0(X, K_X+p^eA)@>>> H^0(p^eQ, K_{p^eQ}+p^e(A-Q))@.\to H^1(X, K_X+p^e(A-Q))\\
@VV{{\operatorname{Tr}}}_{X}^e(A) V @VV \Psi V\\
H^0(X, K_X+A)@>\rho >> H^0(Q, K_Q+A-Q).
\end{CD}$$]{} By Serre duality, we obtain the vanishing $$H^1(X, K_X+p^e(A-Q))=0.$$ On the other hand, $\Psi$ is surjective because $\Psi$ satisfies $$\begin{aligned}
{{\operatorname{Tr}}}_Q^e:H^0(Q, K_Q+p^e(A-Q))&\to &H^0(p^eQ, K_{p^eQ}+p^e(A-Q))\\
&\overset{\Psi}\to& H^0(Q, K_Q+A-Q).\end{aligned}$$ Therefore, the composition map $\rho \circ {{\operatorname{Tr}}}_X^e(A)$ is surjective. Then, we see that ${{\operatorname{Tr}}}_{X}^e(A)$ is a non-zero map by $H^0(Q, K_Q+A-Q)\neq 0$.

\[good-1-form\] In this step, we prove that, if $\deg A=1$, then there exists a point $Q\in X$ such that the natural injective map $$H^0(X, \omega_X(p^eA-p^eQ))\to H^0(X, \omega_X(p^eA-(p^e-1)Q))$$ is not surjective.

Since $H^0(Q, L)\simeq k$ for every invertible sheaf $L$, we obtain the following exact sequence $$\begin{aligned}
0&\to& H^0(X, \omega_X(p^eA-p^eQ))\to H^0(X, \omega_X(p^eA-(p^e-1)Q))\to k\\
&\to& H^1(X, \omega_X(p^eA-p^eQ)).\end{aligned}$$ Therefore, it is sufficient to show $$h^1(X, \omega_X(p^eA-p^eQ))=h^0(X, -(p^eA-p^eQ))=0$$ for some point $Q\in X$. Note that the first equality follows from Serre duality. Assume the contrary, that is, assume that $p^eA\sim p^eQ$ for every point $Q\in X$. Since the genus $g(X)$ is not zero, 