22.1.1 A Locally Lipschitz Selection, Pseudogradients
When you have a functional ϕ defined on a Banach space X, ϕ^{′}
(u)
is in X^{′} and it isn’t
obvious how you can understand it in terms of an element in X like what is done with
Hilbert space using the Riesz representation theorem. However, there is something called
a pseudogradient which is defined next.
Definition 22.1.5Let ϕ : X → ℝ be C^{1}. Then v is a pseudogradient for ϕ at x ifthe following hold.
∥v∥
_{X}≤ 2
∥ϕ′(x )∥
_{X′}
′
∥ϕ (x)∥
_{X′}^{2}≤
′
〈ϕ (x),v〉
A pseudogradient field V is a locally Lipschitz selection of G
(x)
where G
(x)
is definedto be the set of pseudogradients of ϕ at x. Thus V
(x )
∈ G
(x)
and V
(x)
is apseudogradient for ϕ at each x a regular point of ϕ.
Note how this generalizes the case of Hilbert space. In the Hilbert space case, you
have ϕ^{′}
(x)
which technically is in H^{′} and you have the gradient, written here as ∇ϕ
which is in H such that
′
(∇ ϕ(x),v)H ≡ 〈ϕ (x),v〉H′,H
the existence of ∇ϕ
(x)
coming from the Riesz representation theorem which also gives
that ∇ϕ
(x)
= R^{−1}ϕ^{′}
(x)
and so
∥∇ ϕ (x)∥
_{H} =
∥ϕ′(x)∥
_{H′} so the above two conditions
hold for the gradient field except for one thing. Why is x →∇ϕ
(x)
locally Lipschitz. We
don’t know this, but with a pseudogradient field, we do. Also, the pseudogradient field is
only required at regular points of ϕ where ϕ^{′}
(x)
≠0. If you had strict inequalities holding
in the above definition, then they would continue to hold for
xˆ
near x. Thus if you
had
∥v∥X < 2∥ϕ′(x)∥X′ , ∥ϕ′(x)∥2X ′ < 〈ϕ′(x) ,v〉
and Γ
(x)
were the set of such v, then there would be an open set U containing x such
that ∩_{ˆx
∈U}Γ
(ˆx)
≠∅. In fact, the intersection would contain v.
This very nice lemma is from Gasinski L. and Papageorgiou N. [?]. It is
a lovely application of Stone’s theorem and partitions of unity for a metric
space.
Lemma 22.1.6Let Y be a metric space and let X be a normed linear space. (We willwant to add in X.) Let Γ : Y →P
(X )
such that Γ
(y)
is a nonempty convexset. Suppose that for each y ∈ Y, there exists an open set U containing y suchthat
∅ ⁄= ∩ˆy∈UΓ (ˆy)
Then there exists a locally Lipschitz map γ : Y → X such that γ
(y)
∈ Γ
(y)
for ally.
Proof:Let U denote the collection of all open sets U such that the nonempty
intersection described above holds. Let V be a locally finite open refinement which also
covers. Thus for any V ∈V
∅ ⁄= ∩ˆy∈VΓ (ˆy)
because it is a smaller intersection. Let
{ϕV}
_{V ∈V} be a partition of unity subordinate to
the open covering V. In fact, we can have ϕ_{V } locally Lipschitz. This follows from the
above construction of the partition of unity in Theorem 14.1.1. Pick x_{V }∈∩_{ŷ∈V }Γ
(ˆy)
.
Then consider
∑
γ (y) = xV ϕV (y)
V∈V
It is clearly locally Lipschitz because near any point y, it is a finite sum of Lipschitz
functions. Pick y ∈ Y. Then it is in some V ∈V. In fact, it is finitely many, V_{1},
⋅⋅⋅
,V_{n}
and for other V ∈V, ϕ_{V }
(y)
= 0. Therefore,
∑n
γ (y) = xViϕVi (y)
i=1
which is a convex combination of the x_{V i}. Now x_{V i}∈∩_{ŷ∈V i}Γ
(ˆy)
⊆ Γ
(y)
, this for each i.
Hence this is a convex combination of points in a nonempty convex set Γ
(y)
. Thus
γ
(y)
∈ Γ
(y)
. ■
The following lemma says that if ϕ is C^{1} on X, then it has a pseudogradient field on
′
{x : ϕ (x) ⁄= 0}
, the set of regular points.
Lemma 22.1.7Let ϕ be a C^{1}function defined on X a Banach space. Then thereexists a pseudogradient field for ϕ on the set of regular points. (V
(x)
∈ G
(x )
andx → V
(x)
is locally Lipschitz on the set of regular points.)
Proof:First consider whether G
(x)
, the set of pseudogradients of ϕ at x is
nonempty for ϕ^{′}
(x)
≠0. From the definition of the operator norm, there exists u such that
∥u∥
_{X} = 1 and
′
〈ϕ (x),u〉
≥ δ
′
∥ϕ (x)∥
_{X′} where δ ∈
(0,1)
. Then let v = ru
′
∥ϕ (x)∥
_{X′}
where r ∈
(1,2)
.
〈ϕ′(x),v〉 = 〈ϕ′(x),ru ∥ϕ′(x)∥〉 = r 〈ϕ′(x),u〉∥ϕ′(x)∥ ≥ rδ∥ϕ′(x)∥2
Then choose r,δ such that rδ > 1 and r < 2. Then if these were chosen this way in the
above reasoning, it follows that
∥v∥ < 2∥ϕ′(x)∥ and 〈ϕ′(x),v〉 > ∥ϕ′(x)∥2.
That ϕ^{′}
(x)
≠0 is needed to insure that the above strict inequalities hold.
Thus, letting Y be the metric space consisting of the regular points of ϕ,the continuity
of ϕ^{′} implies that the above inequalities persist for all y close enough to x. Thus there is
an open set U containing x such that v satisfies the above inequalities for x replaced with
arbitrary y ∈ U. Thus
v ∈ ∩ G (y)
y∈U
Since it is clear that each G
(y)
is convex, Lemma 22.1.6 implies the existence of a locally
Lipschitz selection from G. That is x → V
(x )
is locally Lipschitz and V
(x)
∈ G
(x)
for
all regular x. ■
It will be important to consider y^{′} = f
(y)
where f is locally Lipschitz and y is just in
a Banach space. This is more complicated than in Hilbert space because of the lack of a
convenient projection map.
Theorem 22.1.8Let f : U → X be locally Lipschitz where X is a Banachspace and U is an open set. Then there exists a unique local solution to theIVP
y′ = f (y), y(0) = y0 ∈ U
Proof:Let B be a closed ball of radius R centered at y_{0} such that f has Lipschitz
constant K on B. Then
∫ t
y1(t) = y0 + f (y0)ds
0
and if y_{n}
(t)
has been obtained,
∫ t
yn+1 (t) = y0 + 0 f (yn (s))ds (22.1.3)
(22.1.3)
Now t < T where T is so small that
∥f (y0)∥
Te^{KT}< R.
Claim:
∥yn(t)− yn− 1(t)∥
≤
∥f (y0)∥
t^{n}K^{n−1}
(n1−1)!
.
Proof of claim:First
∫
t
∥y1(t)− y0∥ ≤ 0 ∥f (y0)∥ ds ≤ ∥f (y0)∥t
Now suppose it is so for n. Then
∫ t
∥yn+1(t)− yn(t)∥ ≤ ∥f (yn(s))− f (yn− 1(s))∥ds
0
Alternate Proof of Theorem 22.1.8:Let B be a closed ball of radius R centered
at y_{0} such that f has Lipschitz constant K on B. Let γ be as in Lemma 22.1.9. Consider
g
(x)
≡ f
(γ(x))
. Then
∥g (x) − g (y)∥ = ∥f (γ(x))− f (γ(y))∥ ≤ K ∥γ (x) − γ(y)∥ ≤ 3K ∥x− y∥ .
Now consider for y ∈ C
([0,T ],X )
∫ t
Fy(t) ≡ y0 + g(y(s))ds
0
Then
∫ t
∥F y(t)− Fz (t)∥ ≤ K ∥y(s)− z(s)∥ds
0
Thus, iterating this inequality, it follows that a large enough power of F is a
contraction map. Therefore, there is a unique fixed point. Now letting y be this fixed
point,
∫
t
∥y(t)− y0∥ ≤ 0 3K ∥y(s)− y0∥ds+ ∥f (y0)∥T
It follows that
∥y(t)− y0∥ ≤ ∥f (y0)∥Te3KT
Choosing T small enough, it follows that
∥y(t)− y0∥
< R on
[0,T ]
and so
γ has no effect. Thus this yields a local solution to the initial value problem.
■
In the case that U = X, the above argument shows that there exists a solution on
some [0,T) where T is maximal.
∫
t
y (t) = y0 + 0 f (y (s))ds, t < T
Suppose T < ∞. Suppose ∫_{0}^{T}
∥f (y(s))∥
ds < ∞. Then you can consider
y_{0} + ∫_{0}^{T}f
(y(s))
ds as an initial condition for the equation and obtain a unique solution
z valid on
[T,T + δ]
. Then one could consider ŷ
(t)
= y
(t)
for t < T and for t ≥ T,ŷ
(t)
= z
(t)
. Then for t ∈
[T,T + δ]
,
∫ T ∫ t
ˆy(t) = z(t) = y0 + 0 f (y(s))d + T f (ˆy(s))ds
∫ T ∫ t
= y0 + f (ˆy(s))d + f (ˆy(s))ds
0 T
and so in fact, for all t ∈
[0,T + δ]
,
∫ t
ˆy (t) = y0 + f (yˆ(s))ds
0
contrary to the maximality of T. Hence it cannot be the case that T < ∞.
Thus it must be the case that ∫_{0}^{T}
∥f (y (s))∥
ds = ∞ if the solution is not
global.
From the above observation, here is a corollary.
Corollary 22.1.10Let f : X → X be locally Lipschitz where X is a Banach space.Then there exists a unique local solution to the IVP
y′ = f (y), y (0) = y0
If f is bounded, then in fact the solutions exists on
[0,T]
for any T > 0.
Proof:Say
∥f (x)∥
≤ M for all M. Then letting [0,
Tˆ
) be the maximal interval, it
must be the case that ∫_{0}^{Tˆ
}
∥f (y(t))∥
dt = ∞, but this does not happen if f is bounded.
■
Note that this conclusion holds just as well if f has linear growth,
∥f (u)∥
≤ a + b
∥u∥
for a,b ≥ 0. One just uses an application of Gronwall’s inequality to verify a similar
conclusion.
One can also give a simple modification of these theorems as follows.
Corollary 22.1.11Suppose f : X → X is continuous and f is locally Lipschitz onU, an open subset of X, a Banach space. Suppose also that f
(x)
= 0 for allx
∕∈
U and that
∥f (x )∥
< M for all x ∈ X. Then there exists a solution to theIVP
y′ = f (y), y (0) = y
0
for t ∈
[0,T]
for any T > 0.
Proof: Let T be given. If y_{0}
∕∈
U, there is nothing to show. The solution is
y
(t)
≡ y_{0}. Suppose then that y_{0}∈ U. Then by Theorem 22.1.8, there exists a
unique solution to the initial value problem on an interval [0,
ˆT
) of maximal
length. If
ˆT
= T, then as t_{n}→ T,
{y(tn)}
must converge. This is because for
t_{m}< t_{n},
∥y(tn)− y(tm)∥ ≤ M |tn − tm|
showing that this is a Cauchy sequence. Since all such sequences lead to a Cauchy
sequence, it must be the case that lim_{t→T}y