= Solution
Take the boundary-fitted grid $h=1/(M+1)$ and assume the standard <consistency of a numerical method> hypothesis $u\in C^4(\overline\Omega)$ with bounded fourth partial <derivatives>. <Taylor expansion> along each coordinate, including a bounded fourth-order remainder, gives the <five-point Laplacian> <truncation error>
$$
\tau_{ij}:=\Delta_hu(ih,jh)-f(ih,jh),\qquad |\tau_{ij}|\le C_0h^2.
$$
The numerical equation and exact grid values therefore imply $A_h\mathbf e=\boldsymbol\tau$, where $A_h=-\Delta_h$ with zero boundary values. Its <discrete sine transform> <eigenvectors> are $\sin(\pi rih)\sin(\pi sjh)$, $1\le r,s\le M$, and its <eigenvalues> are
$$
\lambda_{rs}=\frac4{h^2}\left[\sin^2\frac{\pi rh}2+\sin^2\frac{\pi sh}2\right].
$$
The one-dimensional sine vectors form an <orthogonal basis>, hence so do their <tensor products>. The least <eigenvalue> occurs at $r=s=1$. Since $\sin(\pi h/2)\ge h$ for $0<h\le1$, $\lambda_{11}\ge8$, so the Euclidean <operator norm> of $A_h^{-1}$ is at most $1/8$. There are $M^2$ residual components, and $Mh<1$, giving
$$
\boxed{\|\mathbf e\|_2\le\tfrac18\|\boldsymbol\tau\|_2
\le\tfrac18 C_0h^2M\le\tfrac18 C_0h.}
$$
The factor $M$ matters: pointwise second-order <consistency of a numerical method> becomes first order in the unweighted Euclidean grid norm.
The PDF does not explicitly impose regularity on the exact solution. Without such a hypothesis its unrestricted error claim is false. Here is a counterexample, with a classical solution smooth in the open square and continuous with zero boundary values:
$$
u(x,y)=g(x)\sin(\pi y),\qquad g(x)=\sqrt x(1-x),\qquad f=\Delta u.
$$
The sampled numerical solution separates as $v_i\sin(\pi jh)$. Its error $e_i=v_i-g(ih)$ solves $A e=\tau$, where $A=-\Delta_{h,x}+\lambda_y I$ and $\lambda_y=4h^{-2}\sin^2(\pi h/2)$. The one-dimensional residual is $\tau_i=\Delta_{h,x}g(ih)-g''(ih)+(\pi^2-\lambda_y)g(ih)$. At the first node,
$$
\tau_1=(\sqrt2-7/4)h^{-3/2}-(2\sqrt2-11/4)h^{-1/2}+O(h^2).
$$
The leading coefficient is strictly negative. Moreover $g''''<0$ on $(0,1)$, so the triangular-average formula for the second difference and <concavity> of $g''$ give $\Delta_{h,x}g\le g''$. Thus the positive part of every residual is at most $Ch^2$. The matrix $A$ has a nonnegative inverse, and its first diagonal inverse entry is at least $1/(2h^{-2}+\lambda_y)$, as follows from the nonnegative Neumann expansion of its tridiagonal matrix. The discrete comparison principle bounds the inverse image of the residual's positive part by $Ch^2$, while the negative first residual gives $e_1\le-c\sqrt h+Ch^2$. Hence $|e_1|\ge c'\sqrt h$ for small $h$. Since $\sum_{j=1}^M\sin^2(\pi jh)=1/(2h)$, the full Euclidean error is bounded below by a positive constant. It cannot be $O(h)$. Thus \b[the boxed estimate holds for the intended smooth solution class; a regularity condition is genuinely necessary].
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