Trees and the Agmon Metric
Anna Maltsev
University of Bristol
March 23, 2015
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 1 / 19
Introduction
Main question
How do L2 eigenfunctions decay on a quantum graph?
Prior work:
Liouville and Green for 1D, Agmon: negative energies
Hislop-Post: Worked with “radial trees,” i.e. identical lengths andidentical branching numbers, and a potential added at the vertices asa vertex condition; positive energy localization
Random lengths model
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 2 / 19
Introduction
Main question
How do L2 eigenfunctions decay on a quantum graph?
Prior work:
Liouville and Green for 1D, Agmon: negative energies
Hislop-Post: Worked with “radial trees,” i.e. identical lengths andidentical branching numbers, and a potential added at the vertices asa vertex condition; positive energy localization
Random lengths model
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 2 / 19
Introduction
Main question
How do L2 eigenfunctions decay on a quantum graph?
Prior work:
Liouville and Green for 1D, Agmon: negative energies
Hislop-Post: Worked with “radial trees,” i.e. identical lengths andidentical branching numbers, and a potential added at the vertices asa vertex condition; positive energy localization
Random lengths model
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 2 / 19
Introduction
Agmon philosophy
For negative eigenvalues (actually E < lim inf V ), use tricky integration byparts to show that ψ ∈ L2 and Hψ = Eψ =⇒ exp(ρ(0, x))ψ ∈ L2, whereρ is a metric. Once L2 exponential decay is established, they can be usedto establish pointwise estimates showing exponential decrease.The method works essentially if x : V (x)− E ≤ 0 is compact, and theexpected metric is essentially an ”action integral” of (V (x)− E )1/2+ overthe minimizing path.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 3 / 19
Introduction
Notations:
Our setup:
Quantum graph Γ; edges are segments of the real line
A function space K on Γ so that for φ ∈ K1 φ is twice differentiable on the edges2 φ is continuous at the vertices
3 if v is a vertex, ej are the edges adjacent to v , and φej is φ restrictedto the edge ej , then
∑φ′ej (v) = 0 where the derivative is taken in the
direction away from the vertex v
1 whatever comes in must come out2 Kirchoff condition
4 Notice: if f , g ∈ K then so is fg .Differential operator H = −∆ + V (x) acting on K
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 4 / 19
Introduction
Notations:
Our setup:
Quantum graph Γ; edges are segments of the real line
A function space K on Γ so that for φ ∈ K1 φ is twice differentiable on the edges2 φ is continuous at the vertices3 if v is a vertex, ej are the edges adjacent to v , and φej is φ restricted
to the edge ej , then∑φ′ej (v) = 0 where the derivative is taken in the
direction away from the vertex v
1 whatever comes in must come out2 Kirchoff condition
4 Notice: if f , g ∈ K then so is fg .Differential operator H = −∆ + V (x) acting on K
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 4 / 19
Case Studies The Line
Case study: the Line
Definition 1
Let ρE (x , y) =∫ yx
√(V (t)− E )+dt.
Notes:
In multiple dimensions or on graphs with non-trivial topology, takethe minimum over all possible paths from x to y .
Theorem 2 (Agmon)
Let H = −∆ + V with V real and continuous be a closed operatorbounded below with σ(H) ⊂ R. Suppose E is an eigenvalue of H and thatsupp(E − V (x))+ is compact. Suppose ψ ∈ L2 is an eigenfunction of H.Then for any � > 0 there exists a constant c� such that∫
e2(1−�)ρE (x)|ψ(x)|2dx ≤ c�
adapted from Hislop-Sigal book
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 5 / 19
Case Studies The Line
Case study: the Line
Definition 1
Let ρE (x , y) =∫ yx
√(V (t)− E )+dt.
Notes:
In multiple dimensions or on graphs with non-trivial topology, takethe minimum over all possible paths from x to y .
Theorem 2 (Agmon)
Let H = −∆ + V with V real and continuous be a closed operatorbounded below with σ(H) ⊂ R. Suppose E is an eigenvalue of H and thatsupp(E − V (x))+ is compact. Suppose ψ ∈ L2 is an eigenfunction of H.Then for any � > 0 there exists a constant c� such that∫
e2(1−�)ρE (x)|ψ(x)|2dx ≤ c�
adapted from Hislop-Sigal bookAnna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 5 / 19
Agmon Metric on the Line
Agmon on the Line: Proof ingredient # 1
Lemma 3
For φ ∈ L2 and Fα bounded and satisfying V − E −∣∣∣F ′αFα ∣∣∣2 > δ, we get that
〈Fαφ, (H − E )1
Fαφ〉 ≥ δ‖φ‖2
Proof:〈Fαφ, (H − E )
1
Fαφ
〉=
∫(V − E )|φ|2 + (Fαφ)′
(φ
Fα
)′+ BT
=
∫(V − E )|φ|2 + |φ′|2 −
∣∣∣∣F ′αφFα∣∣∣∣2 ≥ δ‖φ‖2.
Dropped the term |φ′|2, since positive. Integrated by parts, since φ ∈ L2and Fα is bounded. Note δ is independent of the upper bound on Fα.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 6 / 19
Agmon Metric on the Line
Agmon on the Line: Proof ingredient # 1
Lemma 3
For φ ∈ L2 and Fα bounded and satisfying V − E −∣∣∣F ′αFα ∣∣∣2 > δ, we get that
〈Fαφ, (H − E )1
Fαφ〉 ≥ δ‖φ‖2
Proof:〈Fαφ, (H − E )
1
Fαφ
〉=
∫(V − E )|φ|2 + (Fαφ)′
(φ
Fα
)′+ BT
=
∫(V − E )|φ|2 + |φ′|2 −
∣∣∣∣F ′αφFα∣∣∣∣2 ≥ δ‖φ‖2.
Dropped the term |φ′|2, since positive. Integrated by parts, since φ ∈ L2and Fα is bounded. Note δ is independent of the upper bound on Fα.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 6 / 19
Agmon Metric on the Line
Proof Ingredient # 2
Lemma 4
Let η be a smoothed characteristic function of {V − E > δ} such that η′has compact support and ψ be an L2 solution of (H − E )ψ = 0. Then〈
F 2αηψ, (H − E )ηψ〉
=〈η′(ηF 2α
)′ψ,ψ
〉≤ C‖ψ‖2. (3.1)
For the equality can use that (H − E )ψ = 0 and integrate by partsThe inequality follows because η′ has compact support
Can construct such η because we assume E to be an eigenvaluebelow the spectrum, equivalent to V − E ≤ δ compact. Here η = 1on the exterior set and 0 on the compact set
Here we have taken φ from before to be ηFαψ, so thatδ‖ηFαψ‖2 ≤ C‖ψ‖2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 7 / 19
Agmon Metric on the Line
Proof Ingredient # 2
Lemma 4
Let η be a smoothed characteristic function of {V − E > δ} such that η′has compact support and ψ be an L2 solution of (H − E )ψ = 0. Then〈
F 2αηψ, (H − E )ηψ〉
=〈η′(ηF 2α
)′ψ,ψ
〉≤ C‖ψ‖2. (3.1)
For the equality can use that (H − E )ψ = 0 and integrate by partsThe inequality follows because η′ has compact support
Can construct such η because we assume E to be an eigenvaluebelow the spectrum, equivalent to V − E ≤ δ compact. Here η = 1on the exterior set and 0 on the compact set
Here we have taken φ from before to be ηFαψ, so thatδ‖ηFαψ‖2 ≤ C‖ψ‖2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 7 / 19
Agmon Metric on the Line
Proof Ingredient # 2
Lemma 4
Let η be a smoothed characteristic function of {V − E > δ} such that η′has compact support and ψ be an L2 solution of (H − E )ψ = 0. Then〈
F 2αηψ, (H − E )ηψ〉
=〈η′(ηF 2α
)′ψ,ψ
〉≤ C‖ψ‖2. (3.1)
For the equality can use that (H − E )ψ = 0 and integrate by partsThe inequality follows because η′ has compact support
Can construct such η because we assume E to be an eigenvaluebelow the spectrum, equivalent to V − E ≤ δ compact. Here η = 1on the exterior set and 0 on the compact set
Here we have taken φ from before to be ηFαψ, so thatδ‖ηFαψ‖2 ≤ C‖ψ‖2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 7 / 19
Agmon Metric on the Line
Proof Ingredient # 2
Lemma 4
Let η be a smoothed characteristic function of {V − E > δ} such that η′has compact support and ψ be an L2 solution of (H − E )ψ = 0. Then〈
F 2αηψ, (H − E )ηψ〉
=〈η′(ηF 2α
)′ψ,ψ
〉≤ C‖ψ‖2. (3.1)
For the equality can use that (H − E )ψ = 0 and integrate by partsThe inequality follows because η′ has compact support
Can construct such η because we assume E to be an eigenvaluebelow the spectrum, equivalent to V − E ≤ δ compact. Here η = 1on the exterior set and 0 on the compact set
Here we have taken φ from before to be ηFαψ, so thatδ‖ηFαψ‖2 ≤ C‖ψ‖2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 7 / 19
Agmon Metric on the Line
Proof ingredient # 3
We want to take F as large as possible, but still satisfying the constraint
V − E −∣∣∣F ′F ∣∣∣2 > δ, so we construct it by
F (x) = e(1−δ)∫ x
0 (V (t)−E)+ = e(1−δ)ρE (x)
In above lemmas, we use Fα instead where for α > 0
Fα :=F
1 + αF< Cα
We observe that
(i) Fα(x) < F (x),
(ii) |F ′α(x)| < |F ′(x)|,
(iii)∣∣∣F ′α(x)Fα(x) ∣∣∣ < ∣∣∣F ′(x)F (x) ∣∣∣.
Then can take α→ 0, since the constants throughout do not depend on α.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 8 / 19
Agmon Metric on the Line
Proof ingredient # 3
We want to take F as large as possible, but still satisfying the constraint
V − E −∣∣∣F ′F ∣∣∣2 > δ, so we construct it by
F (x) = e(1−δ)∫ x
0 (V (t)−E)+ = e(1−δ)ρE (x)
In above lemmas, we use Fα instead where for α > 0
Fα :=F
1 + αF< Cα
We observe that
(i) Fα(x) < F (x),
(ii) |F ′α(x)| < |F ′(x)|,
(iii)∣∣∣F ′α(x)Fα(x) ∣∣∣ < ∣∣∣F ′(x)F (x) ∣∣∣.
Then can take α→ 0, since the constants throughout do not depend on α.Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 8 / 19
Agmon Metric on the Line
For a quantum tree:
Same argument carries over to rooted trees with few modifications
Integration by parts still works because of the Kirchoff condition
Can show that if ψ ∈ L2(Γ) then e(1−�)ρE (x)ψ ∈ L2, whereρ =
∫ x0 (V (t)− E )
1/2+ and the integral is taken over the unique path
from the root to x
By a standard method can go from L2 estimates to pointwise.
Are we done?
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 9 / 19
Agmon Metric on the Line The Regular Tree
Case study: the regular tree
Regular tree: starts at a root, each edge has length L and at eachvertex splits into b edges, e.g. the Bethe Lattice.
Fix E < 0. Consider H = −∆, i.e. V = 0 and look for a L2 solution.
The problem: Does there exist an L2 solution and does it decay faster
than e−√|E |x , which would be the corresponding decay on the line.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 10 / 19
Agmon Metric on the Line The Regular Tree
Case study: the regular tree
Regular tree: starts at a root, each edge has length L and at eachvertex splits into b edges, e.g. the Bethe Lattice.
Fix E < 0. Consider H = −∆, i.e. V = 0 and look for a L2 solution.The problem: Does there exist an L2 solution and does it decay faster
than e−√|E |x , which would be the corresponding decay on the line.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 10 / 19
Agmon Metric on the Line The Regular Tree
Construction for the Regular Tree
Transfer matrix: T =
(cosh kL sinh kL1b sinh kL
1bcosh kL
)Both eigenvalues λ1, λ2 are real Since detT = 1/b andTrT =
(1 + 1b
)cosh kL > 2/
√b
Since all transfer matrices are equal by construction, the eigenvectorcorresponding to λ1 will give us an initial condition.
The eigenvalue λ1 is given by
λ1 =
(1
2+
1
2b
)cosh kL−
√((1
2+
1
2b
)cosh kL
)2− 1/b
=1(
b2 +
12
)cosh kL +
√((b2 +
12
)cosh kL
)2 − b ≤1
b cosh kL
(3.2)
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 11 / 19
Agmon Metric on the Line The Regular Tree
Construction for the Regular Tree
Transfer matrix: T =
(cosh kL sinh kL1b sinh kL
1bcosh kL
)Both eigenvalues λ1, λ2 are real Since detT = 1/b andTrT =
(1 + 1b
)cosh kL > 2/
√b
Since all transfer matrices are equal by construction, the eigenvectorcorresponding to λ1 will give us an initial condition.
The eigenvalue λ1 is given by
λ1 =
(1
2+
1
2b
)cosh kL−
√((1
2+
1
2b
)cosh kL
)2− 1/b
=1(
b2 +
12
)cosh kL +
√((b2 +
12
)cosh kL
)2 − b ≤1
b cosh kL
(3.2)
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 11 / 19
Agmon Metric on the Line The Regular Tree
the above solution is in L2 for the tree since∫Γ|φ|2 = C
∑n
bnλ2n1 .
If λ1 < α/√b for α < 1 then the above sum converges.
The factor of 1bn makes the pointwise decay faster than the case ofthe line, where the decay is just e−kx , but this comparison is “unfair.”
If a solution is to be in L2 for a tree the 1/√b factor is required for
convergence. However, we have a factor of 1/b instead, which meansthat even if we consider partial integrals, the decay on the tree will befaster than on the line.
Does this hold more generally?
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 12 / 19
Agmon Metric on the Line The Regular Tree
the above solution is in L2 for the tree since∫Γ|φ|2 = C
∑n
bnλ2n1 .
If λ1 < α/√b for α < 1 then the above sum converges.
The factor of 1bn makes the pointwise decay faster than the case ofthe line, where the decay is just e−kx , but this comparison is “unfair.”
If a solution is to be in L2 for a tree the 1/√b factor is required for
convergence. However, we have a factor of 1/b instead, which meansthat even if we consider partial integrals, the decay on the tree will befaster than on the line.
Does this hold more generally?
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 12 / 19
Agmon Metric on the Line The Regular Tree
the above solution is in L2 for the tree since∫Γ|φ|2 = C
∑n
bnλ2n1 .
If λ1 < α/√b for α < 1 then the above sum converges.
The factor of 1bn makes the pointwise decay faster than the case ofthe line, where the decay is just e−kx , but this comparison is “unfair.”
If a solution is to be in L2 for a tree the 1/√b factor is required for
convergence. However, we have a factor of 1/b instead, which meansthat even if we consider partial integrals, the decay on the tree will befaster than on the line.
Does this hold more generally?
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 12 / 19
Our new result
The Agmon Metric for a General Tree
We label the maximum out-degree on the tree to be dmax we will labeleach vertex by a dmaxary string s. We also denote the edge whichterminates at vertex vs by es .
Theorem 5
Suppose ψ ∈ L2 is a solution of H. Suppose at vertex vs , psj is thefraction of the derivative continuing down branch esj . For the path P fromthe root to x , let
ρ(x) =
∫P
(V (t)− E )+ + 2∑v∈P
δV (t) log(1/pv ).
Then if ψ ∈ L2, then e(1−�)ρψ is square integrable on the path P.
Note: e(1−�)ρψ is not in L2(Γ) but sufficient for pointwise bounds.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 13 / 19
Our new result
Key change
For ingredient # 1 inequality V − E −∣∣∣F ′F ∣∣∣2 > δ has to be satisfied.
We sacrifice continuity.
Get boundary terms in integration by parts. Recall that φ = Fαηψ sofor an edge e starting at v1 and ending at v2:∫
eFφ
d2
dx2
(1
Fφ
)=
∫e
d
dx
(F 2ηψ
d
dx(ηψ)
)−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)= F 2ηψ
d
dx(ηψ)
∣∣∣∣v2v1
−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)Want to take F 2 so that its discontinuity cancels with thediscontinuity of ddxψ, so we can increase F
2 by a factor of 1/p.
Technicality: we have to work with Fα so the discontinuity will be inFα then take limα→∞
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 14 / 19
Our new result
Key change
For ingredient # 1 inequality V − E −∣∣∣F ′F ∣∣∣2 > δ has to be satisfied.
We sacrifice continuity.
Get boundary terms in integration by parts. Recall that φ = Fαηψ sofor an edge e starting at v1 and ending at v2:∫
eFφ
d2
dx2
(1
Fφ
)=
∫e
d
dx
(F 2ηψ
d
dx(ηψ)
)−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)= F 2ηψ
d
dx(ηψ)
∣∣∣∣v2v1
−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)Want to take F 2 so that its discontinuity cancels with thediscontinuity of ddxψ, so we can increase F
2 by a factor of 1/p.
Technicality: we have to work with Fα so the discontinuity will be inFα then take limα→∞
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 14 / 19
Our new result
Key change
For ingredient # 1 inequality V − E −∣∣∣F ′F ∣∣∣2 > δ has to be satisfied.
We sacrifice continuity.
Get boundary terms in integration by parts. Recall that φ = Fαηψ sofor an edge e starting at v1 and ending at v2:∫
eFφ
d2
dx2
(1
Fφ
)=
∫e
d
dx
(F 2ηψ
d
dx(ηψ)
)−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)= F 2ηψ
d
dx(ηψ)
∣∣∣∣v2v1
−∫e
d
dx(Fφ)
d
dx
(1
Fφ
)Want to take F 2 so that its discontinuity cancels with thediscontinuity of ddxψ, so we can increase F
2 by a factor of 1/p.
Technicality: we have to work with Fα so the discontinuity will be inFα then take limα→∞
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 14 / 19
Our new result
Case study: the harmonic millipede
We consider a millipede with segments of length L, at each vertex vk ofwhich there dangles an infinitely long leg ek . There is one leg at eachvertex The energy parameter is E = −1, no potential on the main path, soon each bodily edge, ψ′′ = ψ. On the dangling legs the potential isV (x) = (x + 2)2 − 2, where x is the distance from vk , and we see thatV − E > 0 on the legs. The L2 solutions on the legs solveψ′′ = (V (x)− (−1))ψ, which easily gives cst.e−(x+2)2/2. We differentiateand find that ψ′ek (vk) = −2ψ(vk). The transfer matrix for connectingsolutions from one main-body segment to the next is(
cosh L sinh Lsinh L− 2 cosh L cosh L− 2 sinh L
).
The eigenvalues of this are
eL
2
1±√
1−(
2
eL
)2 ≈ e±Lwhen L is large. The solution decays one-dimensionally on the millipede’sbody and superexponentially on the oscillating legs.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 15 / 19
Our new result
Notice:
1 Not L2 if any p = 0, so 1/p’s are ok
2 This requires a lot of information on ψ
Consider instead an averaged function Ψ.
Can use a similar Agmon argument to show that FΨ ∈ L2
Do not need information on the p’s, but only get information
Only get information on ψ in an averaged sense
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 16 / 19
Our new result
Details
Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.
Let Ψ(x) =∑
y :dist(0,y)=x wyψ(x)
wy =∏
v 1/bv where the product is taken over all vertices v on thepath from 0 to y and bv is the out-degree.
Notice:∑
y :dist(0,y)=x wy = 1, so indeed an average
Ψ is continuous everywhere, and Ψ′(v−) = bΨ′(v+) at the vertices
Notice: the p’s are gone!
‖Ψ(x)‖2L2 ≤∫∞
0 (∑
y |wy |2)(∑
y |ψ(y)|2) ≤ ‖ψ(x)‖2L2At each vertex can increase F 2 by a factor of bv , so
e(1−�)∫
(V−E)1/2+ +12
∑n log(bn)δvn (t)dtΨ is also in L2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 17 / 19
Our new result
Details
Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.Let Ψ(x) =
∑y :dist(0,y)=x wyψ(x)
wy =∏
v 1/bv where the product is taken over all vertices v on thepath from 0 to y and bv is the out-degree.
Notice:∑
y :dist(0,y)=x wy = 1, so indeed an average
Ψ is continuous everywhere, and Ψ′(v−) = bΨ′(v+) at the vertices
Notice: the p’s are gone!
‖Ψ(x)‖2L2 ≤∫∞
0 (∑
y |wy |2)(∑
y |ψ(y)|2) ≤ ‖ψ(x)‖2L2At each vertex can increase F 2 by a factor of bv , so
e(1−�)∫
(V−E)1/2+ +12
∑n log(bn)δvn (t)dtΨ is also in L2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 17 / 19
Our new result
Details
Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.Let Ψ(x) =
∑y :dist(0,y)=x wyψ(x)
wy =∏
v 1/bv where the product is taken over all vertices v on thepath from 0 to y and bv is the out-degree.
Notice:∑
y :dist(0,y)=x wy = 1, so indeed an average
Ψ is continuous everywhere, and Ψ′(v−) = bΨ′(v+) at the vertices
Notice: the p’s are gone!
‖Ψ(x)‖2L2 ≤∫∞
0 (∑
y |wy |2)(∑
y |ψ(y)|2) ≤ ‖ψ(x)‖2L2At each vertex can increase F 2 by a factor of bv , so
e(1−�)∫
(V−E)1/2+ +12
∑n log(bn)δvn (t)dtΨ is also in L2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 17 / 19
Our new result
Details
Consider a tree with equal edge lengths and equal branching numbersat each generation. Let ψ ∈ L2 be an eigenfunction.Let Ψ(x) =
∑y :dist(0,y)=x wyψ(x)
wy =∏
v 1/bv where the product is taken over all vertices v on thepath from 0 to y and bv is the out-degree.
Notice:∑
y :dist(0,y)=x wy = 1, so indeed an average
Ψ is continuous everywhere, and Ψ′(v−) = bΨ′(v+) at the vertices
Notice: the p’s are gone!
‖Ψ(x)‖2L2 ≤∫∞
0 (∑
y |wy |2)(∑
y |ψ(y)|2) ≤ ‖ψ(x)‖2L2At each vertex can increase F 2 by a factor of bv , so
e(1−�)∫
(V−E)1/2+ +12
∑n log(bn)δvn (t)dtΨ is also in L2.
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 17 / 19
Work in Progress
Questions to be settled:
1 Is the L2 eigenfunction unique? Analogues of the the limit point -limit circle dichotomy.
2 What if we add leaves and finite subtrees to our regular infinite tree?“Obvious” that this will only improve the decay.
3 What happens in case of a more general graph? What if there arecycles?
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 18 / 19
Thanks
Thank you!
Anna Maltsev (University of Bristol) Trees and the Agmon Metric March 23, 2015 19 / 19
IntroductionCase StudiesThe Line
Agmon Metric on the LineThe Regular Tree
Our new resultWork in ProgressThanks