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Apply the exponential-symbol order criterion for a multistep method, which follows by substituting the bilateral shift operator $e^{h{\partial }_t}$ into the exact-solution residual. Direct expansion givesAll coefficients through degree three vanish. If $\alpha \ne -5$, the fourth-degree coefficient is nonzero, so the normalized local truncation error has exact order three. At $\alpha =-5$, the fourth-degree coefficient vanishes and the fifth-degree coefficient is $1/10$, which is nonzero. ThusThis is the formal order of the full recurrence. The order-four member is not zero-stable, so it is not convergent. All the convergent members from part (a) have order three. At the degenerate value $\alpha =1$, the full recurrence still has formal order three, while cancellation of its common factor gives Backward Euler method of order one; that cancellation changes the method rather than its order calculation.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded everywhere-defined inverse}\}$. Initially define $T$ on finite linear combinations of the orthonormal basis. The bilateral shift operator $S{e}_n=e_{n+1}$ preserves the Hilbert space norm and extends to a unitary operator on $H$; its inverse is $S^{\ast }{e}_n=e_{n-1}$. Thus $T=S+S^{\ast }$ extends uniquely by density, is bounded with norm at most two, and is self-adjoint.

For $v_N=(2N+1{)}^{-1/2}{\sum }_{n=-N}^N{e}_n$, $\Vert v_N\Vert =1$ and the only nonzero coefficients of $(T-2I)v_N$ occur at $-N-1,-N,N,N+1$. HenceIt follows that $\Vert T{v}_N\Vert \to 2$, proving $\boxed{\Vert T\Vert =2}$. Also $\boxed{2\in \sigma (T)}$: a bounded inverse of $T-2I$ would give $1\le \Vert (T-2I{)}^{-1}\Vert \Vert (T-2I)v_N\Vert \to 0$. Thus two belongs to the approximate point spectrum.

An eigenvector $x=\sum x_n{e}_n$ would satisfy $\lambda x_n=x_{n-1}+x_{n+1}$. Self-adjointness and the norm bound force $\lambda \in [-2,2]$. If $|\lambda |<2$, write $\lambda =2\cos \theta$ with $0<\theta <\pi$; the general complex solution of the linear recurrence relation isThe Cesàro mean of $|x_n{|}^2$ over $n=-N,\dots ,N$ tends to $|A{|}^2+|B{|}^2$, since the mixed geometric sum divided by $2N+1$ tends to zero. For an ${\ell }^2$ sequence the same mean tends to zero. Therefore $A=B=0$. At $\lambda =\pm 2$, the general solution is $(A+Bn)(\pm 1{)}^n$, again square summable only when zero. This also covers $\lambda =0$ through $\theta =\pi /2$. Thus $\boxed{T\text{ has no eigenvectors}}$.

Finally the bounded sequence $e_{3j}$ has images $T{e}_{3j}=e_{3j-1}+e_{3j+1}$ with pairwise disjoint supports. Distinct images have distance two, so no subsequence converges. Hence $\boxed{T\text{ is not compact}}$.

For $x\in X$,The map $S$ defined by $(Sx{)}_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a unitary operator.

If $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ for every integer $n$. When $|\lambda |>1$ the positive tail is not square summable, when $|\lambda |<1$ the negative tail is not square summable, and when $|\lambda |=1$ neither tail is square summable unless $x_0=0$. The case $\lambda =0$ also forces every coordinate to vanish. Thus

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has