Polynomial Norms Amir Ali Ahmadi (Princeton University) Georgina Hall Joint work with: Amir Ali Ahmadi (Princeton University) Etienne de Klerk (Tilburg University)

Norms: definition A norm is a function ||⋅||: ℝ 𝑛 →ℝ that satisfies: (1) positivity: 𝑥 ≥0, ∀𝑥∈ ℝ 𝑛 and 𝑥 =0⇒𝑥=0 (2) homogeneity: 𝜆𝑥 = 𝜆 ⋅ 𝑥 , ∀ 𝜆∈ℝ,∀𝑥∈ ℝ 𝑛 (3) triangle inequality: 𝑥+𝑦 ≤ 𝑥 + 𝑦 , ∀𝑥,𝑦∈ ℝ 𝑛 𝑥∈ ℝ 2 𝑥 ∞ =1} NB: 𝑥 ∞ = max 𝑖 𝑥 𝑖 𝑥∈ ℝ 2 𝑥 2 =1} 𝑥 2 = 𝑖 𝑥 𝑖 2 𝑥 1 = 𝑖 | 𝑥 𝑖 | 𝑥∈ ℝ 2 𝑥 1 =1}

When does a polynomial produce a norm? Is this a norm? 𝑝 𝑥 1 , 𝑥 2 =5 𝑥 1 2 −2 𝑥 1 𝑥 2 + 𝑥 2 NO, not homogeneous Is this a norm? 𝑝 𝑥 1 , 𝑥 2 =5 𝑥 1 2 −2 𝑥 1 𝑥 2 + 𝑥 2 2 NO, not 1-homogeneous Necessary condition: has to be the 𝑑 𝑡ℎ root of a degree 𝑑 homogeneous polynomial Is this enough? 1-homogeneity Positivity Triangle inequality  ? ?

Can be checked in polynomial time The quadratic case square root Classic example: 𝑥 2 = 𝑥 1 2 +…+ 𝑥 𝑛 2 Are there others? 𝑓 𝑥 = 𝑥 𝑇 𝑄𝑥 is a norm ⇕ 𝑄≻0 quadratic Can be checked in polynomial time What about higher degree?

Characterizations of polynomial norms (1/3) Theorem 1: If 𝑓 is a form: 𝑓 1/𝑑 is a norm ⇔ 𝑓 is convex and positive definite. Proof: (⇒) Norms are convex and 𝑑 𝑡ℎ power of nonnegative convex function is convex. (⇐)For triangle inequality: let 𝑔= 𝑓 1/𝑑 𝑆 𝑓 = 𝑥 𝑓 𝑥 ≤1 = 𝑥 𝑓 1/𝑑 𝑥 ≤1 ={𝑥|𝑔 𝑥 ≤1}= 𝑆 𝑔 As 𝑓 is convex, 𝑆 𝑓 is convex and so is 𝑆 𝑔 . 𝑥 𝑔(𝑥) , 𝑦 𝑔(𝑦) ∈ 𝑆 𝑔 ⇒𝑔 𝑔 𝑥 𝑔 𝑥 +𝑔 𝑦 ⋅ 𝑥 𝑔 𝑥 + 𝑔 𝑦 𝑔 𝑥 +𝑔 𝑦 ⋅ 𝑦 𝑔 𝑦 ≤1 ⇒𝑔 𝑥+𝑦 𝑔 𝑥 +𝑔(𝑦) ≤1⇒𝑔 𝑥+𝑦 ≤𝑔 𝑥 +𝑔(𝑦)

Characterizations of polynomial norms (2/3) Theorem 2: If 𝑓 is a form: 𝑓 1/𝑑 is a norm ⇔ 𝑓 is strictly convex. Proof: We show that 𝑓 is strictly convex ⇔ 𝑓 is convex and positive definite. (⇒) Strict convexity ⇒ 𝑓 𝑦 >𝑓 𝑥 +𝛻𝑓 𝑥 𝑇 𝑦−𝑥 , ∀𝑦≠𝑥 For 𝑥=0, this becomes 𝑓 𝑦 >0 as 𝑓 𝑥 =0 and 𝛻𝑓 𝑥 =0 (𝑓 is a form). First order characterization of strict convexity

Characterizations of polynomial norms (3/3) ⇐ By contradiction, suppose that 𝑓 is not strictly convex but it is convex and positive definite: ∃𝑥,𝑦 such that 𝑓 𝑥+𝑦 2 = 1 2 𝑓 𝑥 + 1 2 𝑓 𝑦 . Let 𝑔 𝛼 =𝑓 𝑥+𝛼 𝑦−𝑥 : 𝑓 is nonnegative (pd form) ⇒𝑔 is also nonnegative ⇒ 𝑔 is constant. 𝑓 is radially unbounded (pd form) ⇒𝑔 cannot be constant. Has to go through all 3 points 𝒈(𝜶) ?? 𝑔 is not strictly convex Has to be convex but it is convex and univariate ⇒𝒈 is affine. 𝜶 1/2 1

Are all norms polynomial norms? No, the 1-norm ⋅ 1 is a norm but not a polynomial norm for 𝑛>1. But… Theorem: Any norm can be approximated by a polynomial norm arbitrarily well; i.e., for any norm ⋅ , for any 𝜖>0, ∃ an integer 𝑑 and a convex positive definite form 𝑝 of degree 2𝑑 s.t. max 𝑥∈ 𝑆 𝑛−1 | 𝑥 − 𝑝 1 2𝑑 𝑥 |<𝜖. 𝟏−𝝐 𝑩 𝒙 𝒊 𝒗 𝒊 Proof: We show that 1-level set of any norm can be approximated by the 1-level set of some polynomial norm. The result follows by a simple scaling argument. 𝒑 𝒚 = 𝒊 𝒗 𝒊 𝑻 𝒚 𝒗 𝒊 𝑻 𝒙 𝒊 𝟐𝒅 𝑩= 𝒙 𝒙 ≤𝟏}

Complexity results Theorem: Testing whether the 4 𝑡ℎ root of a quartic form is a polynomial norm is NP-hard. Proof [Adapted from a proof by Ahmadi et al.]: Reduction from CLIQUE: NP-hard problem Input: Graph 𝐺=(𝑉,𝐸) and an integer 𝑘 Decision problem: tests whether 𝐺 contains a maximum clique of size >𝑘. 𝜔 𝐺 ≤𝑘⇔ −2𝑘 𝑖,𝑗∈𝐸 𝑥 𝑖 𝑥 𝑗 𝑦 𝑖 𝑦 𝑗 − 1−𝑘 ( 𝑖 𝑥 𝑖 2 )( 𝑖 𝑦 𝑖 2 )+6 𝑛 2 𝑘(∑ 𝑥 𝑖 4 +∑ 𝑦 𝑖 4 +∑ 𝑥 𝑖 2 𝑥 𝑗 2 +∑ 𝑦 𝑖 2 𝑦 𝑗 2 ) is strictly convex

What we have seen so far… For forms 𝑓, 𝑓 1/𝑑 is a norm ⇔ 𝑓 is strictly convex ⇔ 𝑓 is convex and positive definite The 𝑑 𝑡ℎ root of any form of degree 𝑑 that verifies one of the two conditions above is called a polynomial norm. Not all norms are polynomial norms, but they can be approximated by them arbitrarily well. The problem of testing whether a 𝑑 𝑡ℎ root of a degree-d form is already NP-hard when 𝑑=4. How to efficiently test whether the 𝑑 𝑡ℎ root of a degree-𝑑 form is a polynomial norm? How to optimize over set of polynomial norms?

Sum of squares-based relaxations for nonnegativity A polynomial 𝑝 is a sum of squares if there exist polynomials 𝑞 𝑖 s.t. 𝑝 𝑥 = 𝑖 𝑞 𝑖 2 𝑥 . Being a sum of squares is a sufficient condition for nonnegativity. A polynomial 𝑝(𝑥) of degree 2𝑑 is sos if and only if ∃𝑄≽0 such that where 𝑧= 1, x 1 ,…, 𝑥 𝑛 , 𝑥 1 𝑥 2 ,…, 𝑥 𝑛 𝑑 T is the vector of monomials up to degree 𝑑. Optimizing over the set of sos polynomials is a semidefinite program. Sufficient condition but not necessary – we don’t lose that much

Testing for polynomial norms (1/2) Theorem: If 𝑓 is a degree-2𝑑 form: 𝑓 1/2𝑑 is a polynomial norm ⇔ ∃ 𝑐>0, 𝑟∈ℕ, and an sos form 𝑞(𝑥,𝑦) s.t. 𝒒 𝒙,𝒚 𝒚 𝑻 𝛁 𝟐 𝒇 𝒙 𝒚 sos and 𝒇 𝒙 −𝒄 ∑ 𝒙 𝒊 𝟐 𝒅 ∑ 𝒙 𝒊 𝟐 𝒓 sos. Proof: (⇐) 𝒒 𝒙,𝒚 𝒚 𝑻 𝛁 𝟐 𝒇 𝒙 𝒚 sos ⇒ 𝑦 𝑇 𝛻 2 𝑓 𝑥 𝑦≥0, ∀𝑥,𝑦⇒ 𝛻 2 𝑓 𝑥 ≽0,∀𝑥 ⇒𝑓 convex. 𝒇 𝒙 −𝒄 ∑ 𝒙 𝒊 𝟐 𝒅 ∑ 𝒙 𝒊 𝟐 𝒓 sos ⇒ 𝑓 𝑥 ≥𝑐( 𝑖 𝑥 𝑖 2 ),∀𝑥⇒ 𝑓 pd. We have 𝑓 convex + 𝑓 pd ⇒ 𝑓 1/2𝑑 is a polynomial norm. (⇒) Existence results are consequences of results by Artin and Reznick.

Testing for polynomial norms (2/2) Theorem: If 𝑓 is a degree-2𝑑 form: 𝑓 1/2𝑑 is a polynomial norm ⇔ ∃ 𝑐>0, 𝑟∈ℕ, and an sos form 𝑞(𝑥,𝑦) s.t. 𝑞 𝑥,𝑦 𝑦 𝑇 𝛻 2 𝑓 𝑥 𝑦 sos and 𝑓 𝑥 −𝑐 ∑ 𝑥 𝑖 2 𝑑 ∑ 𝑥 𝑖 2 𝑟 sos. Remarks: RHS is an algebraic certificate of LHS, testable via SDP. Covers all polynomial norms. Presence of a free multiplier means that we cannot use this test (to our knowledge) for optimizing over polynomial norms.

Optimizing over polynomial norms (1/2) Theorem: For a form 𝑓: 𝛻 2 𝑓 𝑥 ≻0, ∀𝑥≠0⇒ ∃ 𝑟∈ℕ s.t. ∑ 𝑥 𝑖 2 𝑟 ⋅ 𝑦 𝑇 𝛻 2 𝑓 𝑥 𝑦 is sos. Remarks: Generalizes a result of Reznick on pd forms: 𝑓 𝑥 >0,∀𝑥≠0⇒∃𝑟∈ℕ s.t. 𝑓 𝑥 ⋅ ∑ 𝑥 𝑖 2 𝑟 is sos Proof cannot be obtained directly from Reznick: 𝛻 2 𝑓 𝑥 ≻0⇔ 𝑦 𝑇 𝛻 2 𝑓 𝑥 𝑦>0,∀ 𝑥 = 𝑦 =1 The form 𝑦 𝑇 𝛻 2 𝑓 𝑥 𝑦 is not positive definite. The multiplier ( 𝑖 𝑥 𝑖 2 ) only contains the 𝑥 variables.

Optimizing over polynomial norms (2/2) Corollary: For a form 𝑓: 𝛻 2 𝑓 𝑥 ≻0, ∀𝑥≠0⇔ ∃ 𝑐>0, 𝑟∈ℕ s.t. ∑ 𝑥 𝑖 2 𝑟 ⋅ (𝑦 𝑇 𝛻 2 𝑓 𝑥 −𝑐 ∑ 𝑥 𝑖 2 𝑑 𝑦) is sos. Remarks: Compared to previous theorem, moving from an implication to an equivalence. RHS is algebraic certificate of LHS (testable via SDP). Covers a subset of polynomial norms: 𝛻 2 𝑓 𝑥 ≻0 , ∀𝑥≠0⇒ 𝑓 strictly convex, but converse is not true, e.g., 𝑓 𝑥 1 , 𝑥 2 = 𝑥 1 4 + 𝑥 2 4

An application to the Joint Spectral Radius (1/2) Problem: Given a set of 𝑛×𝑛 matrices 𝑀= 𝐴 1 ,…, 𝐴 𝑚 , when is the switched linear system 𝑥 𝑘+1 = 𝐴 𝜎(𝑘) 𝑥 𝑘 stable? Joint spectral radius (JSR) of 𝑴= 𝑨 𝟏 ,…, 𝑨 𝒎 : 𝜌 𝐴 1 ,…, 𝐴 𝑚 = lim 𝑘→∞ max 𝜎∈ 1,…,𝑚 𝑘 𝐴 𝜎 𝑘 … 𝐴 𝜎 2 𝐴 𝜎 1 1/𝑘 Generalization of spectral radius of one matrix to a family of matrices Theorem: Switched linear system is stable ⇔ 𝜌 𝐴 1 ,…, 𝐴 𝑚 <1 Goal: compute upperbounds on JSR

An application to the Joint Spectral Radius (2/2) For one matrix 𝐴 For a family of matrices 𝐴 1 ,…, 𝐴 𝑚 𝜌 𝐴 <1 ⇔ There exists a contracting quadratic norm, i.e., 𝑉 𝑥 = 𝑥 𝑇 𝑄𝑥 , 𝑄≻0, s.t. 𝑉 𝐴𝑥 <𝑉 𝑥 . 𝜌 𝐴 1 ,…, 𝐴 𝑚 <1 ⇔ There exists a contracting polynomial norm, i.e., 𝑉 𝑥 = 𝑝 1 𝑑 (𝑥), 𝑝 strictly convex form, s.t. 𝑉 𝐴 𝑖 𝑥 <𝑉 𝑥 ,∀𝑥≠0, ∀𝑖=1,…,𝑚 [Ahmadi and Jungers] Remark: Condition testable using SDP.

Summary We studied conditions under which the 𝑑 𝑡ℎ root of a degree 𝑑 form is a norm. Any such function is a polynomial norm. Any norm can be approximated by a polynomial norm. Testing whether the 𝑑 𝑡ℎ root of a degree-𝑑 form is a norm is NP-hard already in the case where 𝑑=4. Using sum of squares, we presented methods for testing for polynomial norms or optimizing over polynomial norms. We gave an application to upperbounding the joint spectral radius of a switched linear system.

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