A gradient estimate for positive functions on graphs

A gradient estimate for positive functions on graphs Yong Lin, Shuang Liu, Yunyan Yang

arXiv:1509.07981v1 [math.DG] 26 Sep 2015

Department of Mathematics, Renmin University of China, Beijing 100872, P. R. China

Abstract We derive a gradient estimate for positive functions, in particular for positive solutions to the heat equation, on finite or locally finite graphs. Unlike the well known Li-Yau estimate, which is based on the maximum principle, our estimate follows from the graph structure of the gradient form and the Laplacian operator. Though our assumption on graphs is slightly stronger than that of Bauer, Horn, Lin, Lippner, Mangoubi, and Yau (J. Differential Geom. 99 (2015) 359405), our estimate can be easily applied to nonlinear differential equations, as well as differential inequalities. As applications, we estimate the greatest lower bound of Cheng’s eigenvalue and an upper bound of the minimal heat kernel, which is recently studied by Bauer, Hua and Yau (Preprint, 2015) by the Li-Yau estimate. Moreover, generalizing an earlier result of Lin and Yau (Math. Res. Lett. 17 (2010) 343-356), we derive a lower bound of nonzero eigenvalues by our gradient estimate. Keywords: gradient estimate, locally finite graph, Harnack inequality, spectral graph 2010 MSC: 58J35

1. Introduction Let G = (V, E) be a finite or locally finite graph, where V denotes the vertex set and E denotes the edge set. For any edge xy ∈ E, we assume its weight w xy > 0. The degree of x ∈ V is defined P as deg(x) = y∼x w xy , here and throughout this paper we write y ∼ x if xy ∈ E. Let µ : V → R be a finite measure. Then the µ-Laplacian (or Laplacian for short) on G is defined as 1 X w xy ( f (y) − f (x)). ∆ f (x) = µ(x) y∼x The associated gradient form reads 2Γ( f, g)(x) =

1 X w xy ( f (y) − f (x))(g(y) − g(x)). µ(x) y∼x

Write Γ( f ) = Γ( f, f ). Denote Dµ = sup x∈V

deg(x) , µ(x)

d=

sup x∈V, xy∈E

µ(x) . w xy

(1)

Email addresses: [email protected] (Yong Lin), [email protected] (Shuang Liu), [email protected] (Yunyan Yang) Preprint submitted to *** September 29, 2015

The main result in this paper is the following gradient estimate. Theorem 1. Let G = (V, E) be a finite or locally finite graph. Suppose that Dµ < +∞,

d < +∞,

(2)

where Dµ and d are defined as in (1). Then for any positive function u : V → R, there holds √ p 2Γ(u) √ ∆u √ ≤ d + dDµ + Dµ . (3) u u Several special cases are listed below:

(i) If u is a positive solution to the differential inequality ∆u − qu ≤ 0 on V, where q : V → R is a function, then there holds √ √ p 2Γ(u) √ − dq ≤ dDµ + Dµ . u

(ii) If u is a positive solution to the differential inequality ∆u − huα ≤ 0, where α ∈ R, and h : V → R is a function, then there holds √ √ p 2Γ(u) √ − dhuα−1 ≤ dDµ + Dµ . u (iii) If u is a positive solution to the differential inequality ∆u − ∂t u ≤ qu, where q : V × R → R is a function, then there holds √ √ p 2Γ(u) √ ∂t u √ − d − dq ≤ dDµ + Dµ . u u

(iv) If u is a positive solution to the differential inequality ∆u − ∂t u + au log u ≤ 0, where a ∈ R is a constant, then there holds √ √ p 2Γ(u) √ ∂t u √ − d − da log u ≤ dDµ + Dµ . u u

Remark 2. For the corresponding partial differential equations on complete Riemannian manifolds, (i) − (iv) were extensively studied, see for examples [11, 12, 10, 15, 14, 16, 17] and the references there in. At least two points can be seen from Theorem 1: One is that (3) is a global estimate; The other is that (3) can be easily applied to nonlinear elliptic or parabolic equations, as well as differential inequalities. We now analyze the assumption (2), whch is equivalent to sup ♯ {y|y ∼ x} < +∞, x∈V

0 < inf

x∈V, y∼x

µ(x) µ(x) ≤ sup < +∞, w xy x∈V, y∼x w xy

(4)

where ♯ {y|y ∼ x} stands for the number of y ∈ V which is adjacent to x. In fact, suppose (2) holds. Then ♯ {y|y ∼ x} ≤ Dµ d and D1µ ≤ µ(x) w xy ≤ d for any y ∼ x. Hence (4) holds. Conversely, if (4) holds, we have P ♯{y|y ∼ x} deg(x) y∼x w xy = ≤ . µ(x) µ(x) inf x∈V, y∼x µ(x) w xy 2

Then (2) follows immediately. If we replace (2) by (4) in Theorem 1, then the gradient estimate (3) would be  r  √ 2Γ(u) √ ∆u √  N N  (5) ≤ b + b  + , u u a ab  where N = supx∈V ♯ {y|y ∼ x}, a = inf x∈V, y∼x assumption of [3] is Dµ < +∞,

µ(x) w xy

and b = sup x∈V, y∼x

Dw = sup x∈V, y∼x

µ(x) w xy .

Note that the essential

deg(x) < +∞. w xy

(6)

√ It is easy to see that (6) is slightly weaker than (2). All gradient estimates in [3] are about u, where u is a positive solution to a parabolic equation on V. Note that for any positive function u 2 √ √ √ 1 X 2Γ( u)(x) = w xy u(y) − u(x) µ(x) y∼x    1/2 X w xy 1/2  1 X √ 4  √      ≤  w xy u(y) − u(x)  µ(x) µ(x) y∼x y∼x 1/2 !1/2    1 X deg(x) 2  ≤ w xy (u(y) − u(x))  µ(x) µ(x) y∼x p p Dµ 2Γ(u)(x). (7) ≤ If u : V × [0, +∞) → R is a positive solution to the parabolic equation ∆u − ∂t u − qu = 0 on V × [0, +∞), we conclude an analog of ([3], Theorem 4.10) by combining (7) with Theorem 1, p √ √ Γ( u) q ∂t u q q Dµ ( Dµ d + 1) − Dµ d √ − Dµ d ≤ . u 2 2 u As an application of Case (iii) of Theorem 1, we state the following Harnack inequality. Theorem 3. Let G = (V, E) be a finite or locally finite graph satisfying (2). Moreover µmax = sup x∈V µ(x) < +∞ and wmin = inf x∈V, y∼x w xy > 0. Assume u : V × (−∞, +∞) → R is a positive solution to the heat inequality ∆u − ∂t u ≤ qu, where q : V × (−∞, +∞) → R is a function. Then for any (x, T 1 ) and (y, T 2), T 1 < T 2 , we have  r  r   Dµ  (dist(x, y))2 dµmax   u(x, T 1 ) ≤ u(y, T 2 ) exp  D +  (T 2 − T 1 ) +   µ d  T2 − T1 wmin ! Z Z ℓ−1 tk+1 tk+1  X  ℓ2  2 (t − tk ) (q(xk+1 , t) − q(xk , t))dt  q(xk , t)dt + + min ,  2  (T − T ) 2 1 tk tk k=0

where the minimum takes over all shortest paths x = x0 , x1 , · · · , xℓ = y connecting x and y, and tk = T 1 + k(T 2 − T 1 )/ℓ, k = 0, 1, · · · , ℓ. In particular, if there exists some constant C0 such that |q(x, t)| ≤ C0 for all (x, t), then for any x, y ∈ V and T 1 < T 2 , there holds   r  r    Dµ 5  (dist(x, y))2 dµmax     + C0  (T 2 − T 1 ) + . (8) u(x, T 1 ) ≤ u(y, T 2) exp  D +     µ d 3 T2 − T1 wmin  3

In a recent work of F. Bauer, B. Hua and S. T. Yau [2], the Li-Yau inequality on graphs, which is due to F. Bauer, P. Horn, Y. Lin, G. Lipper, D. Mangoubi and S. T. Yau [3], is applied to Liouville type theorems and eigenvalue estimates. Moreover a DGG lemma [7, 6, 8] concerning the minimal heat kernel is established on graphs, and it is used together with the Li-Yau inequality to estimate the upper bound of the minimal heat kernel. Our gradient estimate can be used instead of the Li-Yau estimate in [2]. Using Theorem 1, we can estimate the greatest lower bound of the ℓ2 -spectrum known as Cheng’s eigenvalue [4]. Theorem 4. Let G = (V, E) be a locally finite graph satisfying (2) and λ∗ be the lower √ p greatest bound of the ℓ2 -spectrum of the graph Laplacian ∆. Then we have λ∗ ≤ Dµ + Dµ / d. While Theorem 3 can be used to get an analog of ([2], Theorem 1.2).

Theorem 5. Let G = (V, E) be a finite or locally finite graph satisfying (2). Moreover µmax < +∞ and wmin > 0. Let λ∗ be the greatest lower bound of the ℓ2 -spectrum of the graph Laplacian ∆. Given any ǫ > 0, 0 < γ ≤ 1, β > 0. Let Pt (x, y) be the minimum heat kernel of G. Then there exist positive constants C1 (β, γ, Dµ ) and C2 (ǫ, β, γ, Dµ , d, µmax , wmin ) such that for any x, y ∈ V and t ≥ max{βd(x, y), 1}, ( ) √ (dist(x, y))2 exp (−(1 − γ)λ∗ t) Pt (x, y) ≤ q exp C t − C . 2 1 √ √ 4(1 + 2ǫ)t Vol(B x ( t))Vol(By ( t))

Finally we remark that Theorem 1 can also be used to estimate a lower bound of nonzero eigenvalues of the Laplacian on finite connected graphs. Precisely we have an analog of ([13], Theorem 1.8), namely

Theorem 6. Let G = (V, E) be a finite connected graph, Dµ and d be defined as in (1), and D be its diameter. Moreover we assume w x,y = wyx for all y ∼ x and all x ∈ V. Suppose that λ is a nonzero eigenvalue of −∆. Then there holds λ≥ Remark 7. If µ(x) = deg(x) =

1 !. ( q !) Dµ − 1 Dd exp 1 + Dd Dµ + d P

λ≥

y∼x

(9)

w xy , we have Dµ = 1, and whence (9) becomes

1 .    q  Dd exp 1 + Dd 1 + d1 − 1

We refer the reader to [1, 5] for earlier estimates in terms of the volume of the graph G. Let us describe the method. The proof of Theorem 1 is based on the positivity of the average 1 P of u, i.e. µ(x) y∼x w xy u(y), and its relation with ∆u. To prove Theorem 3, we follow [3] and thereby closely follow [12]. While the proof of Theorems 4, 5 and 6 is adapted from [2] and [13] respectively. The remaining part of this paper is organized as follows. In Section 2, we prove the gradient estimate, Theorem 1. In Section 3, we prove the corresponding Harnack inequality, Theorem 3. Finally Theorems 4, 5 and 6 are proved in Section 4. 4

2. Gradient estimate In this section, we prove Theorem 1 by using a very simple method. Proof of Theorem 1. Special cases (i) − (iv) are immediate consequences of (3). Hence it suffices to prove (3). Since u > 0, we have by definition of Γ(u), 2Γ(u)(x) = ≤ =

≤ Noting that

1 X w xy (u(y) − u(x))2 µ(x) y∼x 1 X 1 X w xy u2 (y) + w xy u2 (x) µ(x) y∼x µ(x) y∼x !2 X µ(x) w xy deg(x) u(y) + u2 (x) w µ(x) µ(x) xy y∼x  2 X w xy  d  u(y) + Dµ u2 (x). µ(x) y∼x

X w xy deg(x) u(y) = ∆u(x) + u(x) µ(x) µ(x) y∼x √ and using an elementary inequality a2 + b2 ≤ a + b, ∀a, b ≥ 0, we get p √ X w xy p u(y) + Dµ u(x) 2Γ(u)(x) ≤ d µ(x) y∼x ! √ p deg(x) d ∆u(x) + u(x) + Dµ u(x) = µ(x) √ √ p ≤ d∆u(x) + ( Dµ + dDµ )u(x).

This leads to (3) and thus ends the proof of the theorem.



3. Harnack inequality In this section, following the lines of [3, 12], we prove a Harnack inequality for positive solution to the parabolic inequality ∆u − ∂t u ≤ qu by using (iii) of Theorem 1. Proof of Theorem 3. Let u be a positive solution to the inequality ∆u − ∂t u ≤ qu. By (iii) of Theorem 1, we have p √ Dµ 1 2Γ(u) . (10) − ∂t log u ≤ Dµ + √ + q − √ u d d We distinguish two cases to proceed. Case 1. x ∼ y. 5

For any s ∈ [T 1 , T 2 ], we have by (10) that u(x, T 1 ) u(x, s) u(y, s) = log + log + log u(x, s) u(y, s) u(y, T 2) Z T2 Z s u(x, s) ∂t log u(y, t)dt − ∂t log u(x, t)dt + log = − u(y, s) s T1  p  Z T2 Z s  Dµ   ≤  Dµ + √  (T 2 − T 1 ) + q(y, t)dt q(x, t)dt + d s T1  Z √ Z T2 p 2Γ(u)(y, t)  1  s 2Γ(u)(x, t) dt + dt − √  u(x, t) u(y, t) s d T1 u(x, s) . (11) + log u(y, s)

log u(x, T 1 ) − log u(y, T 2)

We estimate the above terms respectively. Obviously Z s √ 2Γ(u)(x, t) dt ≥ 0. u(x, t) T1

(12)

Since 2Γ(u)(y, t) = ≥

1 X wyz (u(z, t) − u(y, t))2 µ(y) z∼y

wmin (u(x, t) − u(y, t))2 , µmax

we get T2

p

r Z T 2 2Γ(u)(y, t) wmin u(x, t) − 1 dt. dt ≤ − u(y, t) dµmax s u(y, t) s √ Using an elementary inequality log r ≤ |r − 1|, ∀r > 0, we have 1 − √ d

Z

log where

u(x, s) ≤ ψ(x, y, s), u(y, s)

s u(x, s) ψ(x, y, s) = − 1 . u(y, s)

(13)

(14)

Inserting (12), (13), (14) into (11), and using ([3], Lemma 5.3), we obtain r    Dµ   (T 2 − T 1 ) + ψ(x, y, s) log u(x, T 1 ) − log u(y, T 2 ) ≤  Dµ + d  r Z s Z T2 Z T2 wmin − ψ2 (x, y, t)dt + q(y, t)dt q(x, t)dt + dµmax s T1 s r   r  Dµ  dµmax 1   ≤  Dµ +  (T 2 − T 1 ) + d T2 − T1 wmin Z T2 Z T2 1 q(x, t)dt + + (t − T 1 )2 (q(y, t) − q(x, t))dt. (T − T 1 )2 T 1 2 T1 6

Case 2. x is not adjacent to y. Assume dist(x, y) = ℓ. Take a shortest path x = x0 , x1 , · · · , xℓ = y. Let T 1 = t0 < t1 < · · · < tℓ = T 2 , tk = tk−1 + (T 2 − T 1 )/ℓ, k = 1, · · · , ℓ. By the result of Case 1, we have log u(x, T 1 ) − log u(y, T 2) =

ℓ−1 X k=0

log u(xk , tk ) − log u(xk+1 , tk+1 )




r    r ℓ−1  X  Dµ  dµ 1  max   ≤ Dµ +  (tk+1 − tk ) + d tk+1 − tk wmin  k=0 ! Z tk+1 ℓ−1 Z tk+1 X 1 2 (t − tk ) (q(xk+1, t) − q(xk , t))dt q(xk , t)dt + + (tk+1 − tk )2 tk tk k=0 r   r 2  Dµ  dµmax  (T 2 − T 1 ) + ℓ ≤ Dµ + d T2 − T1 wmin ! Z Z ℓ−1 tk+1 tk+1 X ℓ2 2 + q(xk , t)dt + (t − tk ) (q(xk+1 , t) − q(xk , t))dt . (T 2 − T 1 )2 tk tk k=0 Therefore we conclude log u(x, T 1) − log u(y, T 2)

≤

r   r 2  Dµ  dµmax  (T 2 − T 1 ) + (dist(x, y)) Dµ + d (T 2 − T 1 ) wmin + min F (q)(x, y, T 1 , T 2 ),

where F (q)(x, y, T 1, T 2 ) =

ℓ−1 Z X k=0

tk+1

tk

ℓ2 q(xk , t)dt + (T 2 − T 1 )2

Z

tk

tk+1

2

(t − tk ) (q(xk+1, t) − q(xk , t))dt

!

and the minimum takes over all shortest paths connecting x and y. Hence the first assertion of the theorem follows immediately. Moreover, if |q(x, t)| ≤ C0 for all (x, t), then we have ℓ−1 Z X k=0

and

ℓ−1 X k=0

ℓ2 (T 2 − T 1 )2

Z

tk

tk+1

tk

tk+1

q(xk , t)dt ≤ C0 (T 2 − T 1 )

(t − tk )2 (q(xk+1 , t) − q(xk , t))dt ≤

2C0 (T 2 − T 1 ). 3

This gives the desired result and the proof of the theorem is completed.



4. Further applications of the gradient estimate In this section, as applications of Theorem 1, we prove Theorems 4, 5 and 6. For the proof of Theorems 4 and 5, we follow the lines of [2], the essential difference is that we use Theorem 7

1 instead of the Li-Yau estimate [3]. While the proof of Theorem 6 is an adaptation of [13]. For reader’s convenience, we give the details here. Proof of Theorem 4. Let λ∗ be the greatest lower bound of Cheng’s eigenvalues. By a result of S. Haeseler and M. Keller ([9], Theorem 3.1), if λ < λ∗ , then there would be a positive solution u to ∆u = −λu. We conclude from Case (i) of Theorem 1 that √ √ p 2Γ(u) √ + dλ ≤ dDµ + Dµ . u p p Hence λ ≤ Dµ + Dµ /d. Since λ is arbitrary, we obtain λ∗ ≤ Dµ + Dµ /d.  To prove Theorem 5, we need the following DGG lemma on graphs ([2], Theorem 1.1).

Lemma 8. Let Pt (x, y) be the minimal heat kernel of the graph G = (V, E). Then for any β > 0 and 0 < γ ≤ 1, there exists a constant C1 depending only on β, γ and Dµ such that for any subsets B1 , B2 ⊂ G, t ≥ max{βdist(B1 , B2), 1}, ! XX (dist(B1 , B2))2 ∗ p . Pt (x, y)µ(x)µ(y) ≤ e−(1−γ)λ t Vol(B1 )Vol(B2) exp −C1 4t x∈B y∈B 1

2

Proof of Theorem 5. Fix x, y ∈ V, δ > 0, T 1 = t and T 2 = (1 + δ)t. Applying the Harnack inequality, Theorem 3, to the minimal heat kernel Pt (x, y),   r  r   ′ 2   Dµ  dµ (dist(x, x ))   max ′  D + Pt (x, y) ≤ P(1+δ)t (x , y) exp  δt +    µ     d δt wmin    r  r  √   Dµ  1 dµmax    ≤ P(1+δ)t (x′ , y) exp  , ∀x′ ∈ B x ( t).  δt + Dµ +     d δ wmin 

√ Integrating the above inequality on B x ( t) with respect to x′ , we have   r  r  √   Dµ  1 dµmax    X  Vol(B x( t))Pt (x, y) ≤ exp  µ(x′ )P(1+δ)t (x′ , y). D + δt +    µ    d  δ wmin  √ ′

(15)

x ∈B x ( t)

P Note that h(y, s) = x′ ∈B x ( √t) µ(x′ )P s (x′ , y) is also a positive solution to the heat equation. Applying again the Harnack inequality, Theorem 3, to h(y, s) with T 1 = (1 + δ)t and T 2 = (1 + 2δ)t, we have   r  r   √   Dµ  dµ 1   X max  δt + Vol(By ( t))h(y, (1 + δ)t) ≤ exp  µ(y′ )h(y′ , (1 + 2δ)t). Dµ +     d δ wmin  √ y′ ∈By ( t)

This together with (15) implies that    r  r    Dµ  2 dµmax      Pt (x, y) ≤ exp  2 D +  δt +      µ d  δ wmin  X X 1 µ(x′ )µ(y′ )P(1+2δ)t (x′ , y′ ). √ √ Vol(B x( t))Vol(By( t)) x′ ∈B ( √t) y′ ∈B ( √t) y x 8

√ √ 1 Let t ≥ max{βdist(x, y), 1}. Obviously t ≥ 1+2δ max{βdist(B x( t), By( t)), 1}. Let γ, 0 < γ ≤ 1, be fixed. It follows from Lemma 8 that there exists a constant C1 depending only on γ, β and Dµ such that    r  r     Dµ  2 1 dµ    max  δt + D + Pt (x, y) ≤ exp  2   q  µ    √ √    d δ wmin Vol(B x( t))Vol(By( t))   √ √ 2     dist(B x( t), By( t))      ∗ −(1 − γ)λ (1 + 2δ)t − C . (16) exp   1     4(1 + 2δ)t  

√ √ √ √ If dist(x, y) > 2 t, then dist(B x( t), By( t)) ≥ dist(x, y) − 2 t and thus  √ √ 2 dist(B x ( t), By( t)

(dist(x, y))2 1 − . (17) 4(1 + 2δ)t 4(1 + 4δ)t 2δ √ It is easy to see that (17) still holds if dist(x, y) ≤ 2 t. Inserting (17) into (16), we have    r  r    Dµ  1 2 dµmax C1      2 D + + Pt (x, y) ≤ q  δt +   √ exp  √   µ  d  δ wmin 2δ  Vol(B x( t))Vol(By( t)) ) ( (dist(x, y))2 . (18) exp −(1 − γ)λ∗ (1 + 2δ)t − C1 4(1 + 4δ)t √ Note that t ≥ 1. Choosing 2δ = ǫ/ t in (18), we obtain for t ≥ max{βdist(x, y), 1},   r  r    Dµ 4 dµmax C1  1  √   t D ǫ + + ǫ+ Pt (x, y) ≤ q   √ √ exp    µ  d ǫ wmin ǫ  Vol(B x ( t))Vol(By ( t)) ) ( (dist(x, y))2 . exp −(1 − γ)λ∗ t − C1 4(1 + 2ǫ)t q q Dµ dµmax C1 4 Denoting C2 = Dµ ǫ + ǫ +  d ǫ wmin + ǫ , we finish the proof of the theorem. ≥

R Proof of Theorem 6. Note that V ∆udµ = 0 and that if −∆u = λu, then −∆(cu) = λcu for any constant c ∈ R. We can assume −∆u = λu with sup u = 1 and inf u < 0. Take x1 , xℓ ∈ G such that u(x1 ) = sup u = 1, u(xn ) = inf u < 0, x1 x2 · · · xℓ be the shortest path connecting x1 and xℓ , where (xi , xi+1 ) ∈ E. Then ℓ ≤ D. For any β > 1, note that q µ(xi ) 1 P 2 xi y∈E w x y w xi y (u(xi ) − u(y)) µ(xi ) |u(xi ) − u(xi+1 )| i ≤ β − u(xi ) β − u(xi ) √ √ 2Γ(u)(xi ) d . (19) ≤ β − u(xi ) 9

Since β − u > 0 and u ≤ 1, we have by using Theorem 1, p √ 2Γ(β − u) 2Γ(u) = β−u β−u r   √  ∆(β − u) Dµ   d  + Dµ + ≤ β−u d  r   √  λu Dµ   + Dµ + d  = β−u d  r   √  1 Dµ   . λ + Dµ + ≤ d  β−1 d 

This together with (19) implies

ℓ X |u(xi ) − u(xi+1 )| i=1

β − u(xi )

On the other hand, ℓ X |u(xi ) − u(xi+1 )| i=1

β − u(xi )

r    1 Dµ   . ≤ Dd  λ + Dµ + β−1 d  ℓ X

≥

i=1

ℓ X

≥

i=1

|u(xi ) − u(xi+1 )| log 1 + β − u(xi ) log

(20)

!

β − u(xi+1 ) β − u(xi )

β − u(xℓ ) β − u(x1 ) β ≥ log . β−1

= log

(21)

Combining (20) and (21), we have

Choose β such that

1 Dd

r    1 Dµ  β log − Dµ − λ ≥ (β − 1)   . Dd β−1 d 

β log β−1 − Dµ −

λ≥

q

Dµ d

=

1 Dd .

We obtain

1 !. ( q !) Dµ − 1 Dd exp 1 + Dd Dµ + d 

This completes the proof of the theorem.

Acknowledgements. Y. Lin is supported by the National Science Foundation of China (Grant No.11271011). Y. Yang is supported by the National Science Foundation of China (Grant No.11171347 and Grant No. 11471014). 10

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