3 Trace Results

In this section, if we do not specifically mention, we always assume thatX is aK-regular tree with measure and metric as in Section2.1.

Lemma 3.1 Let1≤p <∞. For everyf ∈ L^p(X), we have that Then it follows from Fubini’s Theorem that

exists forν-a.e.ξ ∈∂X and that the trace Trf satisfies the norm estimates.

To show that the limit in Eq. 3.1exists for ν-a.e. ξ ∈ ∂X, it suffices to show that the Forp >1, it follows from the H¨older inequality that

| ˜f^∗(ξ )|^p |f (0)|^p + we obtain by means of Fubini’s theorem that

Here in the above estimates, the notationsI_x andj (x)are the same ones as those we used in Lemma 3.1. Sinceν(Ix)≈ K^−|x| andR₁ <+∞, we further obtain that

Trace and Density Results on Regular Trees

Hence we obtain from estimates Eq.3.3and Eq.3.4thatf˜^∗ is inL^p_ν(∂X) for 1≤ p <

∞, which gives the existence of the limits in Eq.3.1forν-a.e.ξ ∈ ∂X. In particular, since

| ˜f| ≤ ˜f^∗, we have the estimate

∂X

| ˜f|^pdν|f (0)|^p +

X

g_f^pdμ, and hence the norm estimate

˜f_L^p_ν(∂X) |f (0)| +

X

g_f^pdμ

1/p

= f_N˙^1,p_(X).

Since every f ∈ N^1,p(X)is locally absolutely continuous, a direct computation gives the estimate |f (0)| f_N^1,p_(X). Hence we obtain the following result from the above theorem.

Corollary 3.3 Let1 ≤ p < ∞and assume that R_p < +∞. Then the trace Tr inEq.1.1 gives a bounded linear operator Tr :N^1,p(X)→L^p_ν(∂X).

Next, we study non-existence of the traces when Rp = ∞. Before going to the main theorems, we introduce the following lemma.

Lemma 3.4([31]) Let(, d, μ)be aσ-finite metric measure space. Then the following conditions on(, d, μ)are equivalent:

(i) L^p()⊂L^q()for allp, q ∈ (0,∞)withp > q;

(ii) μ() <+∞.

Theorem 3.5 Let 1 ≤ p < ∞ and assume that R_p = +∞. Then there exists a function u∈ ˙N^1,p(X)such that

[0,ξ )limx→ξu(x)= +∞, for all ξ ∈∂X. (3.5) Proof To construct the function u ∈ ˙N^1,p(X) satisfying Eq. 3.5, it suffices to find a nonnegative measurable functiong : [0,∞)→ [0,∞]such that

+∞

0 g(t)λ(t) dt = +∞

_+∞

0 g(t)^pμ(t)K^tdt <+∞. (3.6) Given suchg, we may define the functionuby settingu(0)=0 andu(x)=_|x|

0 g(t)λ(t) dt for anyx ∈ X. Then it follows from the definition of upper gradient thatg_u :X → [0,∞]

defined byg_u(x)=g(|x|)is an upper gradient ofu. Moreover, we obtain that g_u^p_Lp(X) =

X

g_u^pdμ≈ _+∞

0

g(t)^pμ(t)K^tdt <+∞. Hence the condition Eq.3.6impliesu∈ ˙N^1,p(X)and that Eq.3.5holds.

Forp =1, sinceR₁ =_μ(t)K^λ(t)t

L^∞([0,∞)) = ∞, the sets A_k :=

t ∈ [0,∞): λ(t)

μ(t)K^t ≥2^k

, k ∈ N

form a nonincreasing sequence of subsets of[0,∞)and we have

B_kn λ(t) dt <+∞by replacingB_kn with a suitable bounded subset if necessary. Then we defineg by setting

and from the definition ofB_kn that _+∞ it suffices to show the existence of a functionα satisfying

+∞

Remark 3.6 If additionallyμ(X) < ∞, instead of constructing the above increasing func-tion, we may easily modify the construction so as to obtain a piecewise monotone function

Trace and Density Results on Regular Trees

u ∈ N^1,p(X) with values in[0,1] so thatu(x) = 1 when |x| = t_2j and u(x) = 0 when

|x| = t_2j+1, where tk → ∞ as k → ∞. Then this oscillatory function u belongs to N^1,p(X), but has no limit along any geodesic ray. Hence we obtain the following result.

Proposition 3.7 Let 1 ≤ p < ∞ and assume that R_p = +∞. Ifμ(X) < ∞, then there exists a functionu∈N^1,p(X)such thatlim_[0,ξ )x→ξ u(x)does not exist for anyξ ∈∂X.

Remark 3.8 Since our weights only depend on the distance to the root, Theorem 3.2 and Theorem 3.5 boil down to embeddings on the positive real axis. One of the key properties is thatRp <∞if and only if

L^p(R⁺, K^tμ(t) dt) ⊂L¹(R⁺, λ(t) dt) (3.9) where λ and μ are defined on [0,∞) as specified in the introduction. Consequently, if μ(X) < ∞, then R_p < ∞ implies R_q < ∞ whenever 1 ≤ p < q < ∞. However such an implication does not hold true ifμ(X) = ∞, but finiteness ofR_p is still subject to interpolation, i.e. ifRp <∞andRr < ∞thenRq <∞for everyq ∈ [p, r].

The above results give the full answers to the trace results for the homogeneous New-tonian space N˙^1,p(X)and also for the Newtonian space N^1,p(X) when μ(X) < ∞. We continue towards the caseμ(X)= ∞.

Proposition 3.9 Let1≤p <∞and assumeμ(X)=∞. Then for everyf∈L^p(X), we have lim inf

[0,ξ )x→ξ|f (x)| =0, for a.e.ξ ∈ ∂X, (3.10) and hence Trf =0if Trf exists.

Proof Assume that Eq. 3.10 is false. Then there exist a function f ∈ L^p(X) and a set E ⊂∂X withν(E) >0 such that

lim inf

[0,ξ )x→ξ|f (x)| >0, for allξ ∈ E.

Hence for each ξ ∈ E, there exist a constant (ξ ) > 0 and an integer N (ξ ) := N ((ξ )) such that

|f (x)| ≥(ξ ) >0, for allx ∈ [0, ξ ) with |x| ≥N (ξ ).

It follows from Lemma 3.1 that f^p_Lp(X) =

X

|f (x)|^pdμ≈

∂X

[0,ξ )

|f (x)|^pK^{j (x)}dμ(x) dν(ξ )

≥

E

{x∈[0,ξ ):|x|≥N (ξ )}|f (x)|^pK^{j (x)}dμ(x) dν(ξ )

≥

E

{x∈[0,ξ ):|x|≥N (ξ )}(ξ )^pK^{j (x)}dμ(x) dν(ξ )

=

E

(ξ )^p _∞

N (ξ )

K^{j (t)}μ(t) dt dν(ξ ),

wherej (t)is the largest integer such thatj (t) ≤t+1. Sinceμ(X)= ∞andμ ∈L¹_loc(X), for every integerN (ξ ), we have

_∞

N (ξ )

K^{j (t)}μ(t) dt = ∞.

Since(ξ ) >0 for eachξ ∈E andν(E) >0, we obtain that f^p_Lp(X) = +∞,

which contradicts the fact thatf ∈L^p(X). Thus Eq.3.10holds.

If Trf exists, then Tr|f| also exists. It follows from the definition of the trace Eq.1.1 and Eq.3.10that Tr|f| =0. Hence Trf = 0.

Proposition 3.10 AssumeR₁ = +∞. Then there exists a functionu ∈ N^1,1(X)such that lim_{[0,ξ )}x→ξu(x)does not exist, for anyξ ∈∂X.

Proof It follows fromR₁= _μ(t)K^λ(t)t

L^∞([0,∞)) = ∞that the sequence of sets E_k :=

t ∈ [0,∞): λ(t)

μ(t)K^t ≥2^k

satisfies

|E_k|>0 for anyk ∈N. Hence we may choose a sequence{t_k :t_k ∈ [0,∞)}k∈N with

tk → ∞ as k → ∞ and |Ek ∩ [tk−1, tk]|> 0 for any k ∈N. (3.11) Sinceμ∈L¹_loc([0,∞)), we have that for eachk ∈ N,

0<

_tk

t_k−1

μ(t)K^tdt =: M_k <∞.

By the absolute continuity of integral with respect to measure, we may divide the interval [tk−1, tk]into2^kMksubintervals{Ij}j whose interiors are pairwise disjoint such that

2^kM_k j=1

I_j = [t_k−1, t_k] and 0<

I_j

μ(t)K^tdt ≤2^−k. (3.12) Since|E_k ∩ [t_k−1, t_k]| > 0 from Eq. 3.11, we obtain there is at least one subintervalI_k ∈ {I_j}j such that|E_k ∩I_k|>0. Then we define a functiongby setting

g(t)=

₂

Ek∩Ikλ(t) dt, if t ∈E_k ∩I_k, k ∈N;

0, otherwise.

Since λ(t) is always positive and λ ∈ L¹_loc([0,∞)), the above definition is well-defined.

Next we construct the functionu. For anyk ∈ N, since we have t_k

t_k−1

g(t)λ(t) dt =

E_k∩I_k

2λ(t)

E_k∩I_kλ(t) dt dt =2, (3.13) we may apply the same idea of construction as in Remark 3.6 on{x ∈X :t_k−1 ≤ |x| ≤t_k} to obtain a piecewise monotone function u with upper gradient g_u(x) = g(|x|)and with values in [0,1]so thatu(x) = 0 when |x| = t_k−1, t_k andu(x) = 1 when |x| = t_k where tk−1 < t_k < tk. Then the functionuhas no limit along any geodesic ray.

Trace and Density Results on Regular Trees

Thus it remains to show that u ∈ N^1,1(X). We first estimate theL¹-norm of the upper gradientguofu. By the definitions of functiongand ofEk, it follows from estimate Eq.3.13

that Given[0, ξ )satisfying the above inequality, we have

klim→∞

A_k

|f (x(t))|^pK^tμ(t)dt =0.

Sincef is continuous and

A_kK^tμ(t) dt =1 for eachk, there exists atk ∈Ak such that

sinceR^p < ∞. Since_∞

2.⇒3.This implication is trivial.

3.⇒1.Fix p > 1, and suppose that R^p = ∞. Then, for the sequence of subintervals finiteness of the total measure with

kμ(I_k) < ∞ so as to obtain a piecewise monotone functionu ∈ N^1,p(X)with upper gradientg_u(x) =g(|x|),whereg is from above, and so thatuhas no limit along any geodesic ray. This contradicts 3.

Forp = 1, we have R₁ = R1. For p > 1, it is easy to check that R_p < ∞ implies R^p <∞, while the inverse does not hold true. Furthermore, we will show that the finiteness ofR^p will not imply the finiteness ofR_q for any 1≤q <∞.

For simplicity, we consider the special case whereλandμare piecewise constant. More precisely, assume that

λ(t)=λj, μ(t)=μj, fort ∈ [j, j +1), j ∈ N,

Trace and Density Results on Regular Trees

where{λ_j}j∈N and{μ_j}j∈N are two sequences of positive and finite real numbers. Then ds =d λ(z)=λ_jd|z| and d μ(z)=μ_jd|z|, for j ≤ |z|< j +1, j ∈ N. (3.15) We begin with easily checkable conditions.

Lemma 3.12 Let 1 ≤ p < ∞. Let Xbe a K-regular tree with measure and metric as in Eq.3.15. Then the following hold:

(i) R^p <∞if there exists a constantM >0 such that

sup

forp >1, and

Remark 3.13 The conditions in Lemma 3.12 to determine whetherR^p is finite or not are only sufficient conditions but not necessary ones. Towards this:

For (i), pickμ_j =K^−j(1+j )⁻¹,(λ_j)^p =2^−j(1+j )⁻¹. ThenR^p <∞but

Remark 3.15 The above examples show that R^p < ∞ does not guarantee that R_q < ∞ for some 1 ≤ q < ∞ when μ(X) = ∞ and p > 1. Hence the existence of the trace Tr :N^1,p →L^pν(∂X)is not equivalent to the finiteness ofRq for some 1≤q < ∞.

4 Density

In this section, we focus on the density properties of compactly supported functions in N^1,p(X)and in N˙^1,p(X), 1≤ p < ∞. The function1is defined by1(x) = 1 for allx in Xand we abuse the notation by using∇uto denoteg_uif needed for convenience.

Our first result is an analog of the corresponding result for infinite networks [32], also see [24].

Lemma 4.1 Let1≤p <∞and assume thatμ(X) <∞. Then we have that N₀^1,p(X)=N^1,p(X) ⇐⇒ 1∈ N₀^1,p(X).

Proof Since it follows from μ(X) < ∞ that 1 ∈ N^1,p(X), we obtain that N₀^1,p(X) = N^1,p(X)implies1∈ N₀^1,p(X).

Towards the other direction, the hypothesis 1 ∈ N₀^1,p(X)gives a family of compactly supported functions{1_n}n∈NinN^1,p(X)such that1_n →1 inN^1,p(X)asn→ ∞. Recall

Trace and Density Results on Regular Trees

thatX^m := {x ∈ X : |x| < m}for anym ∈ N. Without loss of generality we may assume that1n is nonnegative for anyn∈ Nand that

1_n−1^p_N1,p(X) < 1

4^pμ(X¹), for alln∈N.

We claim that for anyn∈N, there exists a pointx_n withx_n ∈ X¹such that|1−1_n(x_n)|<

1/4. If not, then we have|1−1_n(x)| ≥ ¹₄ for anyx ∈X¹. Hence we obtain that 1−1_n^p_N1,p(X) ≥ 1−1_n^p_N1,p(X¹) ≥ 1

4^pμ(X¹),

which is a contradiction. By the triangle inequality, we have 1−1_n(x_n)≤ |1−1_n(x_n)| < ¹₄, and hence1_n(x_n) > ³₄.

Next, we claim that we may assume1_n(x) > 1/2 for allx ∈ Xⁿ by selecting a subse-quence of{1n}n∈Nif necessary. Assume that this claim is not true. Then there existsN ∈ N such that for any n ∈ N, there exists a point y_n ∈ X^N with1_n(y_n) ≤ 1/2. Hence for any n∈ N, we have found two pointsx_n ∈X¹andy_n ∈ X^N such that|1_n(x_n)−1_n(y_n)| ≥1/4.

Letγ = [x_n, y_n]be the geodesic connectingx_n andy_n. Then

γ

|∇(1_n)|ds ≥1/4 for any n∈ N.

By an argument similar to that for the estimate Eq.2.1and Eq.2.2, we have that there exists a constantC(N, p, λ, μ) >0 such that

X

|∇(1−1_n)|^pdμ=

X

|∇(1_n)|^pdμ≥C(N, p, λ, μ) >0 for any n∈ N, which is a contradiction to1n →1inN^1,p(X).

Thus, from the arguments above, we may assume that there exists a family of compactly supported functions{1_n}n∈N inN^1,p(X)such that

1_n →1inN^1,p(X)asn→ ∞, 1_n(x)≥ ¹₂ for anyx ∈Xⁿ.

We define1¯n :=min{2·1n,1}for alln∈ N. Then the family(1¯n)n∈N satisfies

⎧⎪

⎨

⎪⎩

1¯_n →1inN^1,p(X)asn→ ∞, 1¯_n ≡1 inXⁿ,

1¯_nis a function with compact support.

(4.1)

Given a function uin N^1,p(X), let us show thatu_n1¯_n → uin N^1,p(X) whereu_n(x)is a truncation ofuwith respect toa_n := 1¯_n−1^−1/2_N1,p(X), namely

u_n(x)= u

|u|a_n if|u| ≥a_n u if|u| ≤a_n .

From the basic properties of truncation (see for instance [11, Section 7.1]), we have that

⎧⎪

⎨

⎪⎩

u_n →uinN^1,p(X)asn→ ∞,

|u_n(x)| ≤a_n,

|∇u_n| ≤3|∇u|.

(4.2)

We first show thatu_n1¯_n →uinL^p(X)asn→ ∞. By the triangle inequality, it follows from Eq.4.1and Eq.4.2that

u_n1¯_n −uL^p(X) ≤ u_n1¯_n−u_nL^p(X)+ u_n−uL^p(X)

≤ a_n1¯_n−1N^1,p(X)+ u_n−uL^p(X)

= 1¯_n−1^1/2_N1,p(X)+ un−uL^p(X) →0 as n→ ∞.

Recall that every function inN^1,p(X)is locally absolutely continuous, see Section2.2. By the product rule of locally absolutely continuous functions, we obtain that

|∇(u_n1¯_n−u)| = |∇(u_n1¯_n−u_n +u_n−u)|

≤ |u_n||∇(1¯_n−1)| + |1¯_n−1||∇u_n| + |∇(u_n−u)|

≤ a_n|∇(1¯_n−1)| + |∇u_n|χ_X\Xⁿ + |∇(u_n −u)|.

Hence we obtain from the triangle inequality and Eq.4.2that

∇(u_n1¯_n−u)L^p(X) ≤ a_n∇(¯1_n−1)L^p(X)+ ∇u_nL^p(X\Xⁿ) + ∇(u_n −u)L^p(X)

≤ 1¯_n−1^1/2_N1,p(X)+3∇uL^p(X\Xⁿ)+ ∇(u_n −u)L^p(X), which tends to 0 as n → ∞. Therefore, u_n1¯_n → u in N^1,p(X) as n → ∞. Since the support ofu_n1¯_n is compact, it follows from the definition ofN₀^1,p(X) thatu ∈ N₀^1,p(X), and henceN₀^1,p(X)=N^1,p(X).

Notice that1 ∈ ˙N^1,p(X)no matter ifμ(X) is finite or not. By slightly modifying the previous proof, we obtain the following result.

Corollary 4.2 Let1 ≤p <∞. Then the following statements are equivalent N˙₀^1,p(X)= ˙N^1,p(X) ⇐⇒ 1∈ ˙N₀^1,p(X)

Applying Lemma 4.1, we obtain our first density result.

Proposition 4.3 Let1 ≤ p < ∞and assume thatμ(X) < ∞. Suppose additionally that R_p = ∞. Then we have that

N₀^1,p(X)=N^1,p(X).

Proof It follows from Lemma 4.1 that it suffices to construct a sequence of compactly supportedN^1,p-functions which converges to1inN^1,p(X).

Forp >1 and

Rp = _∞

0

λ

p

p−1(t)μ^1−1p(t)K^1−tpdt = ∞,

we define the family of functions {ϕ_n}n∈N as follows. For eachn ∈ N, let r_n > n be an integer such that _rn

n

λ^p−1p (t)μ^1−p1 (t)K^1−pt dt ≥2ⁿ. (4.3) We setϕ_n(x)=1 for allx ∈Xⁿ,ϕ_n(x)=0 for allx ∈ X\X^rn and

ϕ_n(x)=1− _|x|

n λ^p−1p (t)μ^1−p1 (t)K^1−pt dt r_n

n λ

p

p−1(t)μ^1−1p(t)K^1−tpdt

Trace and Density Results on Regular Trees

for all x ∈ X^rn \Xⁿ. Since λ^p/μ ∈ L^1/(p_loc ⁻¹⁾([0,∞)) and λ,1/μ > 0, then ϕ_n is well-defined. It is easy to check thatϕn is compactly supported.

By the construction ofϕ_n, an easy computation shows that

∇(ϕ_n(x)−1)=0 (4.4)

Thanks to Eq.4.4and Eq.4.5, we obtain the estimate Sincep >1 and Eq.4.3holds, we obtain that

r_n is a nonincreasing sequence of subset of[0,∞)and that we have

|E_k|>0 for any k ∈ N.

We define a sequence{ϕk}of functions by setting ϕ_k(x)=1− 1

It follows directly from the definition ofϕ_k that eachϕ_k has compact support and that

|ϕ_k −1| ≤2χ_X\X^k, |∇(ϕ_k)(x)−1)| ≤ χE_kn(x) λ(x)|Ek_n|. Hence, thanks toμ(X) < ∞and the definition ofE_kn, we obtain that

ϕk −1N^1,1(X) = ϕk −1L¹(X)+ ∇(ϕk −1)L¹(X)

≤ 2χX\X^kL¹(X)+

X

χE_kn(t)

λ(t)|Ek_n|dμ(t)

2μ(X\X^k)+ 1

|E_kn| _kn

k

μ(t)K^t

λ(t) χ_Ekn(t) dt

≤ 2μ(X\X^k)+ 1

2^k →0 ask → ∞. Henceϕk →1inN^1,1(X)ask → ∞.

By using the same construction of the sequence of compactly supportedN^1,p-functions as the one in the above proof, we obtain the following corollary immediately from Corollary 4.2.

Corollary 4.4 Let1≤p <∞. AssumeR_p = ∞. Then we obtain that N˙₀^1,p(X)= ˙N^1,p(X).

Proposition 4.5 Let1 ≤ p < ∞and assume thatμ(X) < ∞. Suppose additionally that R_p <∞. Then we have

N₀^1,p(X)N^1,p(X).

Proof SupposeN₀^1,p(X)= N^1,p(X). Since 1∈ N^1,p(X), it follows that for everyε > 0, there exists a functionu∈N^1,p(X)with compact support such that

1−uN^1,p(X) < ε. (4.6)

Letξ ∈ ∂X be arbitrary, andx_j := x_j(ξ )be the ancestor ofξ with |x_j| = j andx₀ = 0.

Let 0< < ¹₂μ^1/p_L1([0,1]). By repeating the argument in the beginning of Proof of Lemma 4.1 with the change that we replace μ(X¹)/4^p andX¹ by ^p and [0, x₁(ξ )], respectively, we obtain the existence of x_ξ ∈ [0, x₁(ξ )] for which the function u in Eq. 4.6 satisfies

|1−u(x_ξ)| < ¹₂. By the triangle inequality, we have 1− |u(x_ξ)| ≤ |1−u(x_ξ)| < ¹₂, and hence|u(x_ξ)|> ¹₂.

Notice thatuhas compact support. Then for anyξ ∈∂X, we have limn→∞u(xn(ξ ))=0 and that

1

2 < lim

n→∞|u(x_ξ)−u(x_n(ξ ))| ≤

[0,ξ )

g_uds =

[0,ξ )

g_uλ(x)

μ(x)dμ. (4.7)

Trace and Density Results on Regular Trees

By choosing small enough, the above estimate yields a contradiction, and hence N₀^1,1(X)=N^1,1(X).

Forp >1, by Eq.4.7and the H¨older inequality, we have that 1

By choosing small enough, the above estimate gives a contradiction, and hence N₀^1,p(X)=N^1,p(X)forp >1.

Since N₀^1,p(X) ⊂ N^1,p(X)by definition, we obtainN₀^1,p(X) N^1,p(X) for allp ≥ 1.

Corollary 4.6 Letp ≥1and assume thatR_p <∞. Then we have N˙₀^1,p(X)N˙^1,p(X).

Proof SupposeN˙₀^1,p(X) = ˙N^1,p(X). Since 1∈ ˙N^1,p(X), it follows that for every > 0, there exists a functionu∈ ˙N^1,p(X)with compact support such that

1−u_N˙^1,p_(X) < .

Then by the definition of ourN˙^1,p-norm, we have|u(0)−1|< and hence|u(0)|> 1−. Then using a similar argument to the one in the Proof of Proposition 4.5 (replaceu(x_ξ) withu(0)), we obtain a contradiction. The claim follows.

The above results give a full picture for the density properties for homogeneous Newto-nian spacesN˙^1,p(X)and for Newtonian spacesN^1,p(X)whenμ(X) <∞. When the total measure is infinite, the density results for the Newtonian spaceN^1,p(X)are quite different.

Lemma 4.7 LetK = 1, i.e.,Xbe a 1-regular tree and assume thatμ(X) = ∞. Then for anyf ∈ N^1,p(X), there exists a sequence of compactly supported N^1,p-functions{fn}n∈N

such thatf_n →f inN^1,p(X).

Proof Notice that we may compose any f ∈ N^1,p(X) as f = f⁺ − f⁻ where f⁺ = f ·χ_{f≥0} ≥0 andf⁻= −f ·χ_{f≤0} ≥0. Hence we may assume thatf ≥0.

SinceK =1,∂Xcontains only one pointξ₀and there is a unique geodesic ray. It follows from Proposition 3.9 that

lim inf

[0,ξ0)x→ξ₀f (x)=0. (4.8) Denote byxn the vertex ofXwith|xn| =nwhenn∈N. Then it follows from Eq.4.8that

f (x_n)−

[x_n,ξ₀)

g_f ds ≤0, ∀n∈N. (4.9)

We define functionsf_n by setting f_n(x):=

f (x), if|x| ≤n;

max{0, f (xn)−2

[x_n,x]gf ds}, if|x|> n.

Then it is easy to check that f_n ∈ N^1,p(X), since 0 ≤ f_n ≤ f and g_fn ≤ 2gf. Next, we check thatfn is compactly supported. Assume not. Sincefn is non-increasing for |x| > n by definition, we have thatf_n(x) >0 for any|x|> nand hence that

xlim→ξ₀f_n(x)=f (x_n)−2

[x_n,ξ₀)

g_f ds ≥0.

Combining this with Eq.4.9, we conclude that

[x_n,ξ₀)

g_f ds =0.

Thengf = 0 for|x| > nand it follows from Eq.4.8thatf andfn has to be identically 0 for|x| ≥n, which is a contradiction. Hencef_n is compactly supported.

Trace and Density Results on Regular Trees

At last, we estimate theN^1,p-norm off_n −f. By the fact that 0 ≤ f_n ≤f andg_fn ≤ 2gf, we obtain the estimate

f_n −fN^1,p(X) = f_n−fN^1,p(X∩{|x|≥n})

≤ f_nN^1,p_{(X∩{|x|≥n})}+ f_N^1,p_{(X∩{|x|≥n})}

≤ 3fN^1,p(X∩{|x|≥n}) →0 as n→0,

since f ∈ N^1,p(X). Thus {f_n}n∈N is a sequence of compactly supported N^1,p-functions withf_n →f inN^1,p(X), which finishes the proof.

If_∞

0 λ(t) dt = ∞, thenXis complete and unbounded with respect to distanced and it follows by using suitable cutoff functions thatN₀^1,p(X)= N^1,p(X). Our next result shows that this is also the case whenXis bounded and not complete if we assumeμ(X)= ∞. Theorem 4.8 Let1≤p <∞and assume thatμ(X)= ∞. Then we have that

N₀^1,p(X)=N^1,p(X).

Proof IfK = 1, the result follows directly from Lemma 4.7. Hence we assumeK ≥2 in the ensuing proof.

For any f ∈ N^1,p(X), by the same argument as in Lemma 4.7, we may assume that f ≥0. It suffices to construct a sequence {f_n}n∈N of compactly supportedN^1,p-functions such thatf_n →f inN^1,p(X).

For each n ∈ N, we denote by {x_n,j}^K_j=1ⁿ the vertices of n-level, i.e., |x_n,j| = nfor all j = 1,· · · , Kⁿ. For any xn,j, we study the subtreex_n,j which is a subset of Xwith root x_n,j. More precisely,

_xn,j := {x ∈X:x_n,j < x}.

Since every vertex has exactlyK children, we may divide_xn,j intoK subsets, where each subset contains a subtree whose root is a child of x_n,j and an edge connecting this child withx_n,j. We denote by{ⁱ_x

n,j}^K_i=1theseK subsets.

Fixf ∈ N^1,p(X). We first study the function u := f|_xn,j. If f_N^1,p_(xn,j) > 0, we first modify the functionuto a functionvwithv(x)=v(|x|)for anyx ∈_xn,j, i.e., for any x, y ∈_xn,j with|x| = |y|, thenv(x)=v(y). The modification procedure is as follows:

Step 1 Since_xn,j =_K

i=1_xⁱ

n,j, without loss of generality, we may assume

uN^1,p(¹_xn,j) =min{uN^1,p(_xn,jⁱ ) :i =1,2, . . . , K}. (4.10) Then we define a functionu¹by identically copying the minimalN^1,p-energy subtree of u(here isu|¹

xn,j), to the otherk−1 subtreesⁱ_x

n,j,i=2,· · ·K. More precisely, u¹(x):=

u(x), ifx ∈_x¹

n,j; u|¹_xn,j(y)withy ∈_x¹

n,j,|y| = |x|, ifx ∈_xⁱ

n,j. It follows from Eq.4.10that

u¹N^1,p(_xn,j) ≤ uN^1,p(_xn,j).

Then for any x, y ∈ x_n,j ∩ {x ∈ X : n ≤ |x| ≤ n +1} with |x| = |y|, we have u¹(x)= u¹(y).

Step 2 Denote by{x_n+1,t}^K_t=1theKchildren ofx_n,j. We repeat the Step 1 by replacing the functionuand_xn,j withu¹and_xn+1,t, respectively. Here we repeat the Step 1 for allK subtreesx_n+1,t,t = 1,· · · , K. Hence we obtain a functionu² onx_n,j by additionally lettingu²(x)= u¹(x)ifx ∈ _xn,j withn ≤ |x| ≤ n+1. Moreover, it is easy to check that

u²N^1,p(_xn,j) ≤ u¹N^1,p(_xn,j) ≤ uN^1,p(_xn,j)

and thatu²(x)=u²(y)for anyx, y ∈_xn,j ∩ {x ∈ X:n≤ |x| ≤n+2}with|x| = |y|.

Continuing this procedure, we obtain a sequence of functions{u^k}k∈N. We define v = limk→∞u^k. Then we know from induction that

vN^1,p(_xn,j) ≤ uN^1,p(_xn,j) = fN^1,p(_xn,j) (4.11) and thatv(x)=v(y)for anyx, y ∈_xn,j with|x| = |y|.

The value of functionv(x)only depends on the distanced(x_n,j, x). We may regardvas a function on a 1-regular tree with rootxn,j and infinite measure, sinceμ(x_n,j)= ∞. Hence, from the Proof of Lemma 4.7, we are able to choose a compactly supportedN^1,p-function f_n,j on_xn,j with

f_n,j −vN^1,p(_xn,j) ≤3uN^1,p(_xn,j) =3fN^1,p(_xn,j). (4.12) Then it follows from Eq.4.11and Eq.4.12that

f_n,j −fN^1,p(_xn,j) ≤ f_n,j −vN^1,p(_xn,j)+ v−fN^1,p(_xn,j)

≤ f_n,j −vN^1,p(_xn,j)+ vN^1,p(_xn,j)+ fN^1,p(_xn,j)

≤ 5fN^1,p(_xn,j). (4.13)

Iff_N^1,p_(xn,j) =0, thenf =0 on_xn,j and we just definef_xn,j = f|_xn,j. At last, we define a functionf_n by setting

f_n(x):=

f (x), if|x| ≤n; f_n,j(x), ifx ∈_xn,j.

Then it is easy to check that f_n ∈ N^1,p(X)and thatf_n is compactly supported, sincef_n,j are compactly supported for anyj =1,· · · , Kⁿ. It follows from estimate Eq.4.13that

fn −fN^1,p(X) = fn−fN^1,p(X∩{|x|≥n}) =

Kⁿ

j=1

fn −fN^1,p(_xn,j)

=

Kⁿ

j=1

f_n,j −f_N^1,p_(xn,j) ≤5

Kⁿ

j=1

f_N^1,p_(xn,j)

= 5f_N^1,p_{(X∩{|x|≥n})} →0, asn→0,

sincef ∈N^1,p(X). Thus we have found a sequence{f_n}n∈Nof compactly supportedN^1,p -functions withf_n →f inN^1,p(X), which finishes the proof.

Trace and Density Results on Regular Trees

In document Trace Operators and Classification Criteria for Regular Trees (sivua 35-53)