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Noncommutative Cauchy Bound and Noncommutative Montel Bound for Roots of Polynomials

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Abstract

In 1829, Cauchy derived an upper bound for every root of a complex polynomial using the maximum of the absolute values of the coefficients. In 1931, Montel derived an upper bound using the sum of the absolute values of the coefficients. We derive noncommutative versions of the Cauchy and Montel bounds.

Introduction

In 1829, Cauchy derived the following very important upper bound for every root of a complex polynomial using the maximum of the absolute values of the coefficients [1].

Theorem 1.1. [2, 3, 5] (Cauchy Upper Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+zn∈C[z].p(z) = a_0 + a_1 z + \cdots+ a_{n-1}z^{n-1} + z^n \in\mathbb{C}[z].

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣<1+max⁡0≤j≤n−1∣aj∣.|\lambda| < 1 + \max_{0 \le j \le n-1} |a_j|.

Corollary 1.2. [2, 3, 5] (Cauchy Upper Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈C[z],an≠0.p(z) = a_0 + a_1 z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in\mathbb{C}[z], \quad a_n \ne0.

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣<1+max⁡0≤j≤n−1∣ajan∣=1+1∣an∣max⁡0≤j≤n−1∣aj∣.|\lambda| < 1 + \max_{0 \le j \le n-1} \left|\frac{a_j}{a_n}\right| = 1 + \frac{1}{|a_n|}\max_{0 \le j \le n-1}|a_j|.

By a suitable modification, Corollary 1.2 gives the following result.

Theorem 1.3. [2, 3, 5] (Cauchy Lower Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈C[z],a0≠0, an≠0.p(z) = a_0 + a_1 z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in\mathbb{C}[z], \quad a_0 \ne0, \ a_n \ne0.

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣>11+max⁡1≤j≤n∣aja0∣=11+1∣a0∣max⁡1≤j≤n∣aj∣=∣a0∣∣a0∣+max⁡1≤j≤n∣aj∣.\begin{aligned} |\lambda| > \frac{1}{1+\max_{1\le j\le n}\left|\frac{a_j}{a_0}\right|}=\frac{1}{1+\frac{1}{|a_0|}\max_{1\le j\le n}|a_j|} \\ &=\frac{|a_0|}{|a_0|+\max_{1\le j\le n}|a_j|}. \end{aligned}

In 1931, Montel derived another bound for every root of a complex polynomial using the sum of the absolute values of the coefficients [4].

Theorem 1.4. [2, 3, 5] (Montel Upper Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+zn∈C[z].p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n\in\mathbb{C}[z].

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣≤max⁡{1,∑j=0n−1∣aj∣}.|\lambda|\le\max\left\{1,\sum_{j=0}^{n-1}|a_j|\right\}.

Corollary 1.5. [2, 3, 5] (Montel Upper Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈C[z],an≠0.p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+a_nz^n\in\mathbb{C}[z],\qquad a_n\ne0.

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣≤max⁡{1,∑j=0n−1∣ajan∣}=max⁡{1,1∣an∣∑j=0n−1∣aj∣}.|\lambda|\le\max\left\{1,\sum_{j=0}^{n-1}\left|\frac{a_j}{a_n}\right|\right\}=\max\left\{1,\frac{1}{|a_n|}\sum_{j=0}^{n-1}|a_j|\right\}.

Theorem 1.6. [2, 3, 5] (Montel Lower Bound) Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈C[z],a0≠0, an≠0.p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+a_nz^n\in\mathbb{C}[z],\qquad a_0\ne0,\ a_n\ne0.

If λ∈C\lambda\in\mathbb{C} satisfies p(λ)=0p(\lambda)=0, then

∣λ∣≥1max⁡{1,∑j=1n∣aja0∣}=1max⁡{1,1∣a0∣∑j=1n∣aj∣}|\lambda|\ge\frac{1}{\max\left\{1,\sum_{j=1}^{n}\left|\frac{a_j}{a_0}\right|\right\}}=\frac{1}{\max\left\{1,\frac{1}{|a_0|}\sum_{j=1}^{n}|a_j|\right\}}
=∣a0∣max⁡{∣a0∣,∑j=1n∣aj∣}.=\frac{|a_0|}{\max\left\{|a_0|,\sum_{j=1}^{n}|a_j|\right\}}.

We naturally ask for versions of Theorems 1.1, 1.3, 1.4 and 1.6 for polynomials over Banach algebras (particularly, over C*-algebras). In this article, we derive them.

Noncommutative Cauchy and Noncommutative Montel Bounds

Let A\mathcal{A} be a unital Banach algebra and A[z]\mathcal{A}[z] be the set of all polynomials over A\mathcal{A}. We derive a non-commutative version of Theorem 1.1 as follows.

Theorem 2.1. (Noncommutative Cauchy Upper Bound) Let A\mathcal{A} be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+zn∈A[z].p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n\in\mathcal{A}[z].

Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and

∥un∥=∥u∥n.(1)\lVert u^n\rVert=\lVert u\rVert^n. \tag*{(1)}

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Then

∥u∥<1+max⁡0≤j≤n−1∥aj∥.\lVert u\rVert< 1 + \max_{0\leq j\leq n-1}\lVert a_j\rVert.

Proof. Define M:=max⁡0≤j≤n−1∥aj∥M := \max_{0\leq j\leq n-1}\lVert a_j\rVert. Let u∈Au \in A satisfies p(u)=0p(u) = 0 and ∥un∥=∥u∥n\lVert u^n\rVert= \lVert u\rVert^n. We claim that ∥u∥<1+M\lVert u\rVert< 1 + M. Let us suppose that the claim fails. Then

∥u∥≥1+M>1.\lVert u\rVert\geq1 + M > 1.

Since p(u)=0p(u) = 0, we have

un=−(a0+a1u+⋯+an−1un−1).u^n = -(a_0 + a_1u + \cdots+ a_{n-1}u^{n-1}).

By taking the norm and using (1), we get

∥u∥n=∥un∥=∥−(a0+a1u+⋯+an−1un−1)∥=∥a0+a1u+⋯+an−1un−1∥≤∥a0∥+∥a1u∥+⋯+∥an−1un−1∥≤∥a0∥+∥a1∥∥u∥+⋯+∥an−1∥∥un−1∥≤∥a0∥+∥a1∥∥u∥+⋯+∥an−1∥∥u∥n−1≤M(1+∥u∥+⋯+∥u∥n−1)=M∥u∥n−1∥u∥−1.\begin{aligned} \lVert u\rVert^n &= \lVert u^n\rVert= \lVert-(a_0 + a_1u + \cdots+ a_{n-1}u^{n-1})\rVert\\ &= \lVert a_0 + a_1u + \cdots+ a_{n-1}u^{n-1}\rVert\\ &\leq\lVert a_0\rVert+ \lVert a_1u\rVert+ \cdots+ \lVert a_{n-1}u^{n-1}\rVert\\ &\leq\lVert a_0\rVert+ \lVert a_1\rVert\lVert u\rVert+ \cdots+ \lVert a_{n-1}\rVert\lVert u^{n-1}\rVert\\ &\leq\lVert a_0\rVert+ \lVert a_1\rVert\lVert u\rVert+ \cdots+ \lVert a_{n-1}\rVert\lVert u\rVert^{n-1}\\ &\leq M(1 + \lVert u\rVert+ \cdots+ \lVert u\rVert^{n-1})\\ &= M\frac{\lVert u\rVert^n - 1}{\lVert u\rVert- 1}. \end{aligned}

Hence

∥u∥n≤M∥u∥n−1∥u∥−1.\lVert u\rVert^n \leq M\frac{\lVert u\rVert^n - 1}{\lVert u\rVert- 1}.

Rearranging,

∥u∥−1≤M∥u∥n−1∥u∥n=M(1−1∥u∥n)<M.\lVert u\rVert- 1 \leq M\frac{\lVert u\rVert^n - 1}{\lVert u\rVert^n} = M\left(1 - \frac{1}{\lVert u\rVert^n}\right) < M.

Therefore

1+M≤∥u∥<1+M1 + M \leq\lVert u\rVert< 1 + M

which is a contraction. Hence claim holds which completes the proof. □

Corollary 2.2. (Noncommutative Cauchy Upper Bound.) Let AA be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],an is invertible.p(z) = a_0 + a_1z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in A[z], \quad a_n\text{ is invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0 and

∥un∥=∥u∥n.\lVert u^n\rVert= \lVert u\rVert^n.

Then

∥u∥<1+max⁡0≤j≤n−1∥an−1aj∥≤1+∥an−1∥max⁡0≤j≤n−1∥aj∥.\lVert u\rVert< 1 + \max_{0\leq j\leq n-1}\lVert a_n^{-1}a_j\rVert\leq1 + \lVert a_n^{-1}\rVert\max_{0\leq j\leq n-1}\lVert a_j\rVert.

Note that we used condition (1) in the proof of Theorem 2.1. We are unable to derive the result without this assumption. We now derive noncommutative version of Theorem 1.3.

Theorem 2.3. (Noncommutative Cauchy Lower Bound) Let AA be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],a0 and an are invertible.p(z) = a_0 + a_1z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in A[z], \quad a_0 \text{ and } a_n \text{ are invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0, uu is invertible and

∥(u−1)n∥=∥u−1∥n.\lVert(u^{-1})^n \rVert= \lVert u^{-1} \rVert^n.

Then

1∥u−1∥>11+max⁡1≤j≤n∥a0−1aj∥≥11+∥a0−1∥max⁡1≤j≤n∥aj∥.\frac{1}{\lVert u^{-1} \rVert} > \frac{1}{1+\max_{1\le j\le n}\lVert a_0^{-1}a_j\rVert} \ge\frac{1}{1+\lVert a_0^{-1}\rVert\max_{1\le j\le n}\lVert a_j\rVert}.

Proof. Since uu invertible, we have

0=an+an−1u−1+⋯+a1u−(n−1)+a0u−n.0 = a_n + a_{n-1}u^{-1} + \cdots+ a_1u^{-(n-1)} + a_0u^{-n}.

Define v:=u−1v := u^{-1}. Then

0=an+an−1v+⋯+a1vn−1+a0vn.0 = a_n + a_{n-1}v + \cdots+ a_1v^{n-1} + a_0v^n.

Result follows by applying Corollary 2.2 to vv. □

Corollary 2.4. Let AA be a unital C*-algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],an is invertible.p(z) = a_0 + a_1z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in A[z], \quad a_n \text{ is invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0 and uu is normal or self-adjoint. Then

∥u∥<1+max⁡0≤j≤n−1∥an−1aj∥.\lVert u\rVert< 1 + \max_{0\le j\le n-1}\lVert a_n^{-1}a_j\rVert.

Proof. Since uu is a normal element in a C*-algebra, it is known that ∥u2k∥=∥u∥2k\lVert u^{2^k}\rVert= \lVert u\rVert^{2^k} for all k∈Nk \in\mathbb{N} [6] We claim that ∥un∥=∥u∥n\lVert u^n\rVert= \lVert u\rVert^n for all k∈Nk \in\mathbb{N}. Let us suppose that this fails for a nonzero normal element uu. Let n∈Nn \in\mathbb{N} be such that ∥un∥<∥u∥n\lVert u^n\rVert< \lVert u\rVert^n. Choose mm such that n+m=2kn+m=2^k for some k∈Nk \in\mathbb{N}. Then

∥u2k∥=∥un+m∥=∥unum∥≤∥un∥∥um∥<∥u∥n∥u∥m=∥u∥n+m=∥u∥2k,\lVert u^{2^k}\rVert= \lVert u^{n+m}\rVert= \lVert u^n u^m\rVert\le\lVert u^n\rVert\lVert u^m\rVert< \lVert u\rVert^n\lVert u\rVert^m = \lVert u\rVert^{n+m} = \lVert u\rVert^{2^k},

which is a contradiction. □

Following the proof of Theorem 2.1 we easily get following results.

Theorem 2.5. (Noncommutative Cauchy Upper Bound) Let AA be a unital Banach algebra. Let

p(z)=a0+za1+⋯+zn−1an−1+znan∈A[z],an is invertible.p(z) = a_0 + za_1 + \cdots+ z^{n-1}a_{n-1} + z^na_n \in A[z], \quad a_n \text{ is invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0 and

∥un∥=∥u∥n.\lVert u^n\rVert= \lVert u\rVert^n.

Then

∥u∥<1+max⁡0≤j≤n−1∥ajan−1∥≤1+∥an−1∥max⁡0≤j≤n−1∥aj∥.\lVert u\rVert< 1 + \max_{0\le j\le n-1}\lVert a_ja_n^{-1}\rVert\le1+\lVert a_n^{-1}\rVert\max_{0\le j\le n-1}\lVert a_j\rVert.

Theorem 2.6. (Noncommutative Cauchy Lower Bound) Let AA be a unital Banach algebra. Let

p(z)=a0+za1+⋯+zn−1an−1+znan∈A[z],a0 and an are invertible.p(z) = a_0 + za_1 + \cdots+ z^{n-1}a_{n-1} + z^na_n \in A[z], \quad a_0 \text{ and } a_n \text{ are invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0, uu is invertible and

∥(u−1)n∥=∥u−1∥n.\lVert(u^{-1})^n\rVert= \lVert u^{-1}\rVert^n.

Then

1∥u−1∥>11+max⁡1≤j≤n∥aja0−1∥≥11+∥a0−1∥max⁡1≤j≤n∥aj∥.\frac{1}{\lVert u^{-1}\rVert} > \frac{1}{1+\max_{1\le j\le n}\lVert a_j a_0^{-1}\rVert} \ge\frac{1}{1+\lVert a_0^{-1}\rVert\max_{1\le j\le n}\lVert a_j\rVert}.

Corollary 2.7. Let A\mathcal{A} be a unital C∗C^*-algebra. Let

p(z)=a0+a1z+⋯+zn−1an−1+znan∈A[z],p(z)=a_0+a_1z+\cdots+z^{n-1}a_{n-1}+z^n a_n\in\mathcal{A}[z],

ana_n is invertible. Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and uu is normal or self-adjoint. Then

∥u∥<1+max⁡0≤j≤n−1∥ajan−1∥≤1+∥an−1∥max⁡0≤j≤n−1∥aj∥.\lVert u\rVert<1+\max_{0\le j\le n-1}\lVert a_j a_n^{-1}\rVert\le1+\lVert a_n^{-1}\rVert\max_{0\le j\le n-1}\lVert a_j\rVert.

We now derive noncommutative version of Theorem 1.4.

Theorem 2.8. (Noncommutative Montel Upper Bound) Let A\mathcal{A} be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+zn∈A[z].p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n\in\mathcal{A}[z].

Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and

∥un∥=∥u∥n.(2)\lVert u^n\rVert=\lVert u\rVert^n. \tag*{(2)}

Then

∥u∥≤max⁡{1,∑j=0n−1∥aj∥}.\lVert u\rVert\le\max\left\{1,\sum_{j=0}^{n-1}\lVert a_j\rVert\right\}.

Proof. Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and ∥un∥=∥u∥n\lVert u^n\rVert=\lVert u\rVert^n. If ∥u∥≤1\lVert u\rVert\le1, then the conclusion holds. We therefore assume that ∥u∥>1\lVert u\rVert>1. Since p(u)=0p(u)=0, we have

un=−(a0+a1u+⋯+an−1un−1).u^n=-(a_0+a_1u+\cdots+a_{n-1}u^{n-1}).

By taking norm and using (2), we get

∥u∥n=∥un∥=∥−(a0+a1u+⋯+an−1un−1)∥=∥a0+a1u+⋯+an−1un−1∥≤∥a0∥+∥a1u∥+⋯+∥an−1un−1∥≤∥a0∥+∥a1∥∥u∥+⋯+∥an−1∥∥un−1∥≤∥a0∥+∥a1∥∥u∥+⋯+∥an−1∥∥u∥n−1≤∥a0∥∥u∥n−1+∥a1∥∥u∥n−1+⋯+∥an−1∥∥u∥n−1=∥u∥n−1∑j=0n−1∥aj∥.\begin{aligned} \lVert u\rVert^n&=\lVert u^n\rVert=\lVert-(a_0+a_1u+\cdots+a_{n-1}u^{n-1})\rVert\\ &=\lVert a_0+a_1u+\cdots+a_{n-1}u^{n-1}\rVert\\ &\le\lVert a_0\rVert+\lVert a_1u\rVert+\cdots+\lVert a_{n-1}u^{n-1}\rVert\\ &\le\lVert a_0\rVert+\lVert a_1\rVert\lVert u\rVert+\cdots+\lVert a_{n-1}\rVert\lVert u^{n-1}\rVert\\ &\le\lVert a_0\rVert+\lVert a_1\rVert\lVert u\rVert+\cdots+\lVert a_{n-1}\rVert\lVert u\rVert^{n-1}\\ &\le\lVert a_0\rVert\lVert u\rVert^{n-1}+\lVert a_1\rVert\lVert u\rVert^{n-1}+\cdots+\lVert a_{n-1}\rVert\lVert u\rVert^{n-1}\\ &=\lVert u\rVert^{n-1}\sum_{j=0}^{n-1}\lVert a_j\rVert. \end{aligned}

Therefore

∥u∥n≤∥u∥n−1∑j=0n−1∥aj∥.\lVert u\rVert^n\le\lVert u\rVert^{n-1}\sum_{j=0}^{n-1}\lVert a_j\rVert.

Cancellation of ∥u∥n−1\lVert u\rVert^{n-1} gives the result. □

Note that we used (2) in the proof of Theorem 2.8. We are unable to derive without it.

Corollary 2.9. Let AA be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],an is invertible.p(z) = a_0 + a_1z + \cdots+ a_{n-1}z^{n-1} + a_nz^n \in A[z], \qquad a_n \text{ is invertible}.

Let u∈Au \in A be such that p(u)=0p(u) = 0 and

∥un∥=∥u∥n.\lVert u^n\rVert= \lVert u\rVert^n.

Then

∥u∥<max⁡{1,∑j=0n−1∥an−1aj∥}≤max⁡{1,∥an−1∥∑j=0n−1∥aj∥}.\lVert u\rVert< \max\left\{1,\sum_{j=0}^{n-1}\lVert a_n^{-1}a_j\rVert\right\} \le\max\left\{1,\lVert a_n^{-1}\rVert\sum_{j=0}^{n-1}\lVert a_j\rVert\right\}.

We now derive another noncommutative Montel upper bound with different assumption.

Theorem 2.10. (Noncommutative Montel Upper Bound) Let AA be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+zn∈A[z].p(z) = a_0 + a_1z + \cdots+ a_{n-1}z^{n-1} + z^n \in A[z].

Let u∈Au \in A be such that p(u)=0p(u) = 0 and uu is invertible. Then

1∥u−1∥≤max⁡{1,∑j=0n−1∥aj∥}.\frac{1}{\lVert u^{-1}\rVert} \le\max\left\{1,\sum_{j=0}^{n-1}\lVert a_j\rVert\right\}.

Proof. Let u∈Au \in A satisfies p(u)=0p(u) = 0 and is invertible. If

1∥u−1∥≤1,\frac{1}{\lVert u^{-1}\rVert} \le1,

then the conclusion holds. Let us therefore assume that

1∥u−1∥>1.\frac{1}{\lVert u^{-1}\rVert} > 1.

Now we need to show that

1∥u−1∥≤∑j=0n−1∥aj∥.\frac{1}{\lVert u^{-1}\rVert} \le\sum_{j=0}^{n-1}\lVert a_j\rVert.

Since p(u)=0p(u) = 0 and uu is invertible, we have

1=−(an−1u−1+⋯+a1u−(n−1)+a0u−n).1 = -(a_{n-1}u^{-1} + \cdots+ a_1u^{-(n-1)} + a_0u^{-n}).

By taking the norm, we get

1=∥−(an−1u−1+⋯+a1u−(n−1)+a0u−n)∥=∥an−1u−1+⋯+a1u−(n−1)+a0u−n∥≤∥an−1u−1∥+⋯+∥a1u−(n−1)∥+∥a0u−n∥≤∥an−1∥∥u−1∥+⋯+∥a1∥∥u−(n−1)∥+∥a0∥∥u−n∥≤∥an−1∥∥u−1∥+⋯+∥a1∥∥u−1∥n−1+∥a0∥∥u−1∥n≤∥an−1∥∥u−1∥+⋯+∥a1∥∥u−1∥+∥a0∥∥u−1∥=∥u−1∥(∑j=0n−1∥aj∥).\begin{aligned} 1 &= \lVert-(a_{n-1}u^{-1} + \cdots+ a_1u^{-(n-1)} + a_0u^{-n})\rVert\\ &= \lVert a_{n-1}u^{-1} + \cdots+ a_1u^{-(n-1)} + a_0u^{-n}\rVert\\ &\le\lVert a_{n-1}u^{-1}\rVert+ \cdots+ \lVert a_1u^{-(n-1)}\rVert+ \lVert a_0u^{-n}\rVert\\ &\le\lVert a_{n-1}\rVert\lVert u^{-1}\rVert+ \cdots+ \lVert a_1\rVert\lVert u^{-(n-1)}\rVert+ \lVert a_0\rVert\lVert u^{-n}\rVert\\ &\le\lVert a_{n-1}\rVert\lVert u^{-1}\rVert+ \cdots+ \lVert a_1\rVert\lVert u^{-1}\rVert^{n-1} + \lVert a_0\rVert\lVert u^{-1}\rVert^n \\ &\le\lVert a_{n-1}\rVert\lVert u^{-1}\rVert+ \cdots+ \lVert a_1\rVert\lVert u^{-1}\rVert+ \lVert a_0\rVert\lVert u^{-1}\rVert\\ &= \lVert u^{-1}\rVert\left(\sum_{j=0}^{n-1}\lVert a_j\rVert\right). \end{aligned}

Therefore

1∥u−1∥≤∑j=0n−1∥aj∥.\frac{1}{\lVert u^{-1}\rVert} \le\sum_{j=0}^{n-1} \lVert a_j\rVert.

Note that we used the invertibility of uu in the proof of Theorem 2.10. We are unable to derive the result without this assumption.

Corollary 2.11. (Noncommutative Montel Upper Bound) Let A\mathcal{A} be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],an is invertible.p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+a_nz^n\in\mathcal{A}[z], \qquad a_n\text{ is invertible}.

Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and uu is invertible. Then

1∥u−1∥≤max⁡{1,∑j=0n−1∥an−1aj∥}≤max⁡{1,∥an−1∥(∑j=0n−1∥aj∥)}.\frac{1}{\lVert u^{-1}\rVert}\le\max\left\{1,\sum_{j=0}^{n-1}\lVert a_n^{-1}a_j\rVert\right\}\le\max\left\{1,\lVert a_n^{-1}\rVert\left(\sum_{j=0}^{n-1}\lVert a_j\rVert\right)\right\}.

By doing a similar procedure as in the proof of noncommutative Cauchy lower bounds, we get following result.

Theorem 2.12. (Noncommutative Montel Lower Bound) Let A\mathcal{A} be a unital Banach algebra. Let

p(z)=a0+a1z+⋯+an−1zn−1+anzn∈A[z],a0 and an are invertible.p(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+a_nz^n\in\mathcal{A}[z], \qquad a_0\text{ and }a_n\text{ are invertible}.

Let u∈Au\in\mathcal{A} be such that p(u)=0p(u)=0 and uu is invertible. Then

∥u∥≥1max⁡{1,∑j=1n∥a0−1aj∥}≥1max⁡{1,∥a0−1∥(∑j=1n∥aj∥)}.\lVert u\rVert\ge\frac{1}{\max\left\{1,\sum_{j=1}^{n}\lVert a_0^{-1}a_j\rVert\right\}}\ge\frac{1}{\max\left\{1,\lVert a_0^{-1}\rVert\left(\sum_{j=1}^{n}\lVert a_j\rVert\right)\right\}}.

Conclusions

  1. In 1829, Cauchy derived an upper bound for the zeros of complex polynomials using the maximum of the absolute values of the coefficients [1].

  2. In 1931, Montel derived an upper bound for the zeros of complex polynomials using the sum of the absolute values of the coefficients [4].

  3. In this article, we derived noncommutative versions of Cauchy and Montel bounds.

References

  1. [1]A. L. Cauchy. Sur la resolution des equations numeriques et sur la theorie de l’élimination. Oeuvres, 9(2):87–161, 1829.
  2. [2]Holly P. Hirst and Wade T. Macey. Bounding the roots of polynomials. The College Mathematics Journal, 28(4):292–295, 1997.DOI
  3. [3]Morris Marden. Geometry of polynomials. 2nd ed., volume 3 of Math. Surv. American Mathematical Society (AMS), Providence, RI, 1966.
  4. [4]P. Montel. Sur la limite supérieure des modules des zéros des polynomes. C. R. Acad. Sci., Paris, 193:974–976, 1931.
  5. [5]Q. I. Rahman and G. Schmeisser. Analytic theory of polynomials, volume 26 of Lond. Math. Soc. Monogr., New Ser. Oxford: Oxford University Press, 2002.
  6. [6]M. Takesaki. Theory of operator algebras I., volume 124 of Encycl. Math. Sci. Berlin: Springer, 2002.

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