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Convergence of a numerical method on $[0,T]$ means that its grid error obeys ${\max }_{0\le nh\le T}|y_n-y(nh)|\to 0$ as $h\to 0$, with consistent initial values.

First, the implicit step of Backward Euler method is well-defined for $hL<1$: the map $z\mapsto y_n+hf(t_{n+1},z)$ is a contraction mapping on $\mathbb{R}$, so the Banach fixed-point theorem gives a unique solution. Let $M={\max }_{[0,T]}|y^{\prime \prime }|$, finite under the stated smoothness. The exact solution's step defect isso $|d_n|\le M{h}^2/2$. Subtract the exact and numerical step equations. The Lipschitz condition givesWith exact initial data, sum this geometric progression to obtainFor $hL\le 1/2$ and $nh\le T$, $-\log (1-hL)\le 2hL$, so the bracket is at most $e^{2LT}-1$. Thus the global error is $O(h)$, uniformly on $[0,T]$, and Backward Euler method converges with order one. A vanishing initial error adds only $(1-hL{)}^{-n}|e_0|$.

Order one is sharp: for $y^{\prime }=y$, $y(0)=1$ and $nh=T$, $y_n=(1-h{)}^{-n}=e^T(1+Th/2+O(h^2))$.

For the Dahlquist test equation $y^{\prime }=\lambda y$, the stability function is $R(z)=1/(1-z)$, $z=h\lambda$. Therefore its linear stability domain isThe strict inequality gives decay; equality gives bounded amplification. The whole closed left half-plane lies in this region, so Backward Euler method is A-stable.

Write $e(t)=\exp (2\pi it)$. A sequence in the circle group is an equidistributed sequence if, for every interval $I\subseteq \mathbb{R}/\mathbb{Z}$,where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. Trigonometric polynomials approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant additive characters have integral zero. These facts give the Weyl criterion:This is the link between equidistribution and cancellation in exponential sums.

Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne 0$ and set $z_n=e(m\varphi (n))$. For every fixed $h\ge 1$, the Weyl criterion would giveThe omitted final $h$ terms change a normalized average by at most $h/N$.

Here is the needed Van der Corput inequality for finite scalar sequences. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to $H{\sum }_n{z}_n={\sum }_t{\sum }_{j=0}^{H-1}{z}_{t-j}$ and expand the squared inner sum. Taking $N\to \infty$ first leaves a bound $1/H$; then let $H\to \infty$. Every nonzero Fourier average of $\varphi$ vanishes, so the Weyl criterion makes $\varphi$ equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive $h$, hence for some $h\ne 0$ as requested.

For $\varphi (n)=\sqrt{2}{n}^2$, the difference is ${\Delta }_h\varphi (n)=-2h\sqrt{2}n-h^2\sqrt{2}$. For any nonzero integer $m$, its exponential sum is a constant phase times a geometric progression with ratio $e(-2mh\sqrt{2})\ne 1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt{2})|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves