0.1. Solution of Problem 1. Step 1 Let x ∈ U be a point where the maximum
of |Du| is achieved. If |Du|(x) = 0 there is nothing to prove. Otherwise set ν =
Du(x)/|Du|(x) and consider the function v = ν · Du. v is harmonic in U and C 1 on
U . Therefore, by the maximum principle, there is a y ∈ ∂U such that
|Du|(x) = v(x) ≤ v(y) = ν · Du(y) ≤ |Du|(y) .
We thus conclude that
max |Du| ≤ max |Du| .
∂U
U
(0.1)
Step 2 Without loss of generality we assume that |Du| achieves its maximum in
0 ∈ ∂U and that U ⊂ {x1 ≥ 0}. Let ν := Du/|Du|(0) and e1 := (1, 0, . . . , 0). If
ν · e1 = 0, then |Du|(0) = |Dτ u|(0) = |Dτ g|(0) ≤ |Dg(0)|, where Dτ f denotes the
tangential derivative of f on ∂U . We therefore assume ν · e1 6= 0. Moreover, since −u
and −g satisfy the assumptions of the exercise, we can also assume ν · e1 > 0. Consider
next the points xε := εν. For ε small enough we have xε ∈ U and hence
|Du(0)| = lim
ε↓0
u(xε ) − u(0)
.
ε
(0.2)
Step 3 Let M := maxU |∆g| and d := diam U and consider the function
w(x) := g(x) +
M
dM
x1 − x21
2
2
Observe that ∆w = ∆g − M ≤ 0. Thus ∆w ≤ 0. Moreover, if x ∈ U , then 0 ≤ x1 ≤
|x| ≤ diam (U ) = d. We therefore conclude that
M x1
(d − x1 ) ≥ g(x)
2
Therefore, w ≥ g = u on ∂U and by the maximum principle
w(x) = g(x) +
u(x) ≤ w(x)
for all x ∈ U .
(0.3)
We can therefore use (0.2) and (0.3) to estimate
w(xε ) − u(0)
g(εν)) + dM εν · e1 /2 − M ε2 (ν · e1 )2 /2 − g(0)
= lim
ε↓0
ε↓0
ε
ε
g(εν) − g(0) dM
dM
= lim
+
ν · e1 = Dg(0) · ν +
ν · e1
ε↓0
ε
2
2
dM
≤ |Dg|(0) +
.
(0.4)
2
Recalling that |Du|(0) = max∂U |Du|, d = diam (U ) and M = maxU |∆g|, (0.1) and
(0.4) give the desired inequality.
|Du|(0) ≤ lim
1
2
0.2. Solution of Problem 2. First of all, by the density of Cc∞ (Rn ) in W N,p (Rn ), it
suffices to show the inequality under the additional assumption u ∈ Cc∞ (Rn ).
Step 1 We prove the inequality in the special case n = 1, N = 2 and j = 1. Indeed
we will show that
ku0 kLp ≤ 2kukLp + ku00 kLp
for every u ∈ Cc∞ (R).
Having fixed u ∈ Cc∞ (R) and x ∈ R we have
ˆ
ˆ x+1
ˆ x+1
0
0
0
0
u (x) −
|u (s) − u (x)| ds ≤
u (s) ds ≤
x
x
ˆ
x+1
x+1
s
|u00 (τ )| dτ ds
x
ˆ
x+1
|u00 (τ )| dτ .
|u00 (τ )| dτ ds =
x
(0.6)
x
x
ˆ
x+1
1
|u00 (x+τ )| dτ . (0.7)
00
|u (τ )| dτ ≤ |u(x+1)|+|u(x)|+
0
x
x
ˆ
x
ˆ
≤
We therefore conclude
ˆ x1
ˆ
0
0
|u (x)| ≤ u (s)ds +
x+1
(0.5)
Integrating this last inequality we obtain
p 1/p
ˆ ˆ 1
00
0
dx
|u (x + τ )| dτ
ku kLp ≤ 2kukLp +
R
ˆ ˆ
0
1/p
1
00
p
|u (x + τ )| dτ dx
≤ 2kukLp +
R
≤ 2kukLp + ku00 kLp . (0.8)
0
Step 2 Assume n ∈ N and u ∈ Cc∞ (Rn ). From the previous step we conclude
ˆ
ˆ
p
k∂x1 ukLp dx =
|∂x1 u(x1 , x0 )|p dx1 dx0
n−1
R
R
ˆ
ˆ
ˆ
(0.5)
2
p
0 p
0 p
p 2
|u(x1 , x )| dx1 + |∂x1 u(x1 , x )| dx1 dx0
≤
Rn−1
≤
C̄kukpLp
R
R
+
C̄k∂x21 ukpLp
.
(0.9)
Summing (0.9) over i and taking the p-th rooth we achieve
kDukLp ≤ CkukLp + CkD2 ukLp .
(0.10)
Next, fix any ε ∈]0, +∞[ and define v(x) := ε−n+1 u(x/ε). Note that
ˆ x p dx 1/p
kDukLp =
Du
n
ε
ε
(0.10)
= kDvkLp ≤ C̄kvkLp + C̄kD2 vkLp
= C̄ε1/p kukLp +
C̄
kD2 ukLp .
ε1/p
(0.11)
3
Moreover, from (0.11), for any fixed N , n and p we conclude the existence of a constant
C̃ = C̃(N, n, p) such that
kDN ukLp ≤ C̃δkDN −1 ukLp +
C̃
kDN +1 ukLp
δ
∀u ∈ Cc∞ (Rn ), ∀δ ∈]0, +∞[ . (0.12)
Step 3 Having fixed n and p we now prove the desired inequality by induction on
N . First of all observe that (0.10) provides the starting point of the induction, i.e. the
case N = 2 of the desired inequality. Next, we fix N > 1 and assume the existence of
a constant CN such that
kDj ukLp ≤ CN kukLp + CN kDN ukLp
∀j ∈ {1, . . . , N − 1}, ∀u ∈ Cc∞ (Rn ) . (0.13)
We want to show the existence of a constant CN +1 such that
kDj ukLp ≤ CN +1 kukLp +CN +1 kDN +1 ukLp
∀j ∈ {1, . . . , N }, ∀u ∈ Cc∞ (Rn ) . (0.14)
Observe that (0.14) follows from (0.13) and
kDN ukLp ≤ CkukLp + CkDN +1 ukLp
∀u ∈ Cc∞ (Rn ) .
(0.15)
To show (0.15) we fix δ > 0 (whose choice will be specified later) and observe
kDN ukLp
(0.12)
≤
(0.13)
≤
C̃δkDN −1 ukLp +
C̃
kDN +1 ukLp
δ
C̃CN δkukLp + C̃CN δkDN ukLp +
C̃
kDN +1 ukLp .
δ
At this point choose δ = (2C̃CN )−1 to infer
1
1
kDN ukLp ≤ kukLp + kDN ukLp + CkDN +1 ukLp ,
2
2
from which (0.15) follows easily.
(0.16)