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

2. 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 𝑥 ∞ =1}
NB:
𝑥 ∞ = max 𝑖 𝑥 𝑖
𝑥∈ ℝ 𝑥 2 =1}
𝑥 2 = 𝑖 𝑥 𝑖 2
𝑥 1 = 𝑖 | 𝑥 𝑖 |
𝑥∈ ℝ 𝑥 1 =1}

3. 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  ? ?

4. 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?

5. 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)

6. 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

7. 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 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

8. 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}

9. 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

10. 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?

11. 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

12. 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.

13. 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.

14. 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.

15. 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 𝑥 2 4

16. 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 /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

17. 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.

18. 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.

19. Thank you for listening
Questions?
Want to learn more?