[proofplan] We prove the polynomial assertion by isolating the leading term of $R$ along the negative real axis. After factoring out $x^n$, the lower-order terms become negative powers of $x$ and hence tend to $0$ as $x\to-\infty$. Therefore the leading coefficient dominates, so $|R(x)|$ grows beyond $1$ for some negative real number $x$. The final statement follows directly from the definition of A-stability, which requires $|R(z)|\leq 1$ throughout the left half-plane. [/proofplan]

[step:Factor the polynomial along the negative real axis]Since $R$ is a nonconstant polynomial, there exist $n\in\mathbb{N}$ with $n\geq 1$ and coefficients $a_0,\dots,a_n\in\mathbb{C}$ with $a_n\neq 0$ such that, for every $z\in\mathbb{C}$, \begin{align*} R(z)=\sum_{j=0}^{n} a_j z^j. \end{align*} For every real number $x<0$, factoring out $x^n$ gives \begin{align*} R(x)=x^n\left(a_n+\sum_{j=0}^{n-1} a_j x^{j-n}\right). \end{align*} Here each exponent $j-n$ is a negative integer for $0\leq j\leq n-1$.[/step]

[guided]The point of this factorisation is to separate the unique highest-degree term from all lower-order terms. Since $R$ is nonconstant, it has a positive degree $n\geq 1$, and we may write \begin{align*} R(z)=\sum_{j=0}^{n} a_j z^j \end{align*} for coefficients $a_0,\dots,a_n\in\mathbb{C}$ with leading coefficient $a_n\neq 0$. Now fix a real number $x<0$. The expression $x^n$ is nonzero, so it can be factored from every term: \begin{align*} R(x)=x^n\left(a_n+\sum_{j=0}^{n-1} a_j x^{j-n}\right). \end{align*} The reason this is useful is that the lower-degree terms have become negative powers of $x$. For every $j\in\{0,\dots,n-1\}$, the integer $j-n$ is strictly negative, so $x^{j-n}$ tends to $0$ as $x\to-\infty$ along the real axis. Thus the parenthetical factor will approach the nonzero number $a_n$.[/guided]

[step:Choose a negative point where the leading coefficient dominates]Define the error function \begin{align*} E:(-\infty,0)&\to\mathbb{C} \end{align*} by \begin{align*} E(x)=\sum_{j=0}^{n-1} a_j x^{j-n}. \end{align*} Since each function $x\mapsto x^{j-n}$ tends to $0$ as $x\to-\infty$, the finite sum $E(x)$ tends to $0$ as $x\to-\infty$. Hence there exists $M_1>0$ such that, whenever $x<-M_1$, \begin{align*} |E(x)|\leq \frac{|a_n|}{2}. \end{align*} For such $x$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |a_n+E(x)|\geq |a_n|-|E(x)|\geq \frac{|a_n|}{2}. \end{align*} Therefore, for every $x<-M_1$, \begin{align*} |R(x)|=|x|^n |a_n+E(x)|\geq |x|^n\frac{|a_n|}{2}. \end{align*}[/step]

[guided]We now make precise the statement that the lower-order terms are negligible. Define \begin{align*} E:(-\infty,0)&\to\mathbb{C} \end{align*} by \begin{align*} E(x)=\sum_{j=0}^{n-1} a_j x^{j-n}. \end{align*} For every index $j\in\{0,\dots,n-1\}$, the exponent $j-n$ is negative. Thus $x^{j-n}=1/x^{n-j}$, and since $n-j\geq 1$, this term tends to $0$ as $x\to-\infty$. A finite sum of functions tending to $0$ also tends to $0$, so $E(x)\to 0$ as $x\to-\infty$. Because $a_n\neq 0$, the number $|a_n|/2$ is positive. By the definition of the limit $E(x)\to 0$, there exists $M_1>0$ such that $x<-M_1$ implies \begin{align*} |E(x)|\leq \frac{|a_n|}{2}. \end{align*} For these negative [real numbers](/page/Real%20Numbers) $x$, the reverse triangle inequality gives \begin{align*} |a_n+E(x)|\geq |a_n|-|E(x)|. \end{align*} Using the chosen bound on $E(x)$, we obtain \begin{align*} |a_n+E(x)|\geq |a_n|-\frac{|a_n|}{2}=\frac{|a_n|}{2}. \end{align*} Substituting this estimate into the factorisation \begin{align*} R(x)=x^n(a_n+E(x)) \end{align*} and using multiplicativity of the complex modulus yields \begin{align*} |R(x)|=|x|^n |a_n+E(x)|\geq |x|^n\frac{|a_n|}{2}. \end{align*} This is the domination step: once $x$ is sufficiently far to the left, the leading coefficient cannot be cancelled by the lower-order terms.[/guided]

[step:Send the leading term past the stability bound]Choose $M_2>0$ such that $M_2^n>2/|a_n|$. Let $x\in\mathbb{R}$ satisfy \begin{align*} x<-\max\{M_1,M_2\}. \end{align*} Then $x<0$, $x<-M_1$, and $|x|>M_2$. Hence \begin{align*} |R(x)|\geq |x|^n\frac{|a_n|}{2}>M_2^n\frac{|a_n|}{2}>1. \end{align*} Thus there exists $x\in(-\infty,0)$ such that $|R(x)|>1$.[/step]

[guided]We now choose one negative real number that satisfies both requirements already obtained. The first requirement is $x<-M_1$, which ensures that the lower-order error $E(x)$ is small enough for the leading coefficient estimate. The second requirement is $|x|>M_2$, which makes the factor $|x|^n$ large enough to force the modulus past $1$. Choose $M_2>0$ such that \begin{align*} M_2^n>2/|a_n|. \end{align*} This is possible because $n\geq 1$ and $|a_n|>0$. Now choose $x\in\mathbb{R}$ with \begin{align*} x<-\max\{M_1,M_2\}. \end{align*} Then $x<0$, so $x\in(-\infty,0)$. Also $x<-M_1$, so the domination estimate from the previous step applies. Finally, since $x<-M_2<0$, we have $|x|=-x>M_2$. Therefore \begin{align*} |R(x)|\geq |x|^n\frac{|a_n|}{2}>M_2^n\frac{|a_n|}{2}>1. \end{align*} This proves the promised existence statement: there is a negative real number $x\in(-\infty,0)$ such that $|R(x)|>1$.[/guided]

[step:Conclude that polynomial stability functions cannot be A-stable]Let a one-step method have stability function $R:\mathbb{C}\to\mathbb{C}$, and suppose $R$ is a nonconstant polynomial. Write $\operatorname{Re}(z)$ for the real part of a complex number $z\in\mathbb{C}$. By the result just proved, there exists $x\in(-\infty,0)$ such that $|R(x)|>1$. Since $\operatorname{Re}(x)=x<0$, this point lies in the left half-plane. By the definition of A-stability stated in the theorem, an A-stable method must satisfy $|R(z)|\leq 1$ for every $z\in\mathbb{C}$ with $\operatorname{Re}(z)\leq 0$. Applying this requirement to the negative real point $z=x$ would give $|R(x)|\leq 1$, contradicting $|R(x)|>1$. Therefore no one-step method with a nonconstant polynomial stability function is A-stable.[/step]

[guided]Let a one-step method have stability function $R:\mathbb{C}\to\mathbb{C}$, and suppose $R$ is a nonconstant polynomial. Write $\operatorname{Re}(z)$ for the real part of a complex number $z\in\mathbb{C}$. The polynomial result already proved gives a real number $x\in(-\infty,0)$ such that \begin{align*} |R(x)|>1. \end{align*} Because $x$ is real, its real part is itself, so \begin{align*} \operatorname{Re}(x)=x<0. \end{align*} Thus the complex number $x$ lies in the left half-plane. A-stability, as stated in the theorem, requires the stability bound \begin{align*} |R(z)|\leq 1 \end{align*} for every $z\in\mathbb{C}$ with $\operatorname{Re}(z)\leq 0$. Applying this condition to the particular left-half-plane point $z=x$ would imply \begin{align*} |R(x)|\leq 1. \end{align*} This contradicts the strict inequality $|R(x)|>1$ obtained from the polynomial growth argument. Therefore the assumed one-step method cannot be A-stable, and no one-step method with a nonconstant polynomial stability function is A-stable.[/guided]