16.5 The Tychonoff And Schauder Fixed Point Theorems
First we give a proof of the Schauder fixed point theorem which is an infinite dimensional
generalization of the Brouwer fixed point theorem. This is a theorem which lives in
Banach space. After this, we give a generalization to locally convex topological vector
spaces where the theorem is sometimes called Tychonoff’s theorem. First here is an
interesting example [?].
Exercise 16.5.1Let B be the closed unit ball in a separable Hilbert space H whichis infinite dimensional. Then there exists continuous f : B → B which has no fixedpoint.
Let
{ek}
_{k=1}^{∞} be a complete orthonormal set in H. Let L ∈ℒ
(H, H )
be defined as
follows. Le_{k} = e_{k+1} and then extend linearly. Then in particular,
( )
∑ ∑
L xiei = xiei+1
i i
Then it is clear that L preserves norms and so it is linear and continuous. Note how this
would not work at all if the Hilbert space were finite dimensional. Then define
f
= 1. But then you would need to have x = Lx which is not
so because if x is in the closure of the span of
{ei}
_{i=m}^{∞}, such that the first
nonzero Fourier coefficient is the m^{th}, then Lx is in the closure of the span of
{ei}
_{i=m+1}^{∞}.
This shows you need something other than continuity if you want to get a fixed
point. This also shows that there is a retraction of B onto ∂B in any infinite
dimensional separable Hilbert space. You get it the usual way. Take the line
from x to f
(x)
and the retraction will be the function which gives the point
on ∂B which is obtained by extending this line till it hits the boundary of B.
Thus for Hilbert spaces, those which have ∂B a retraction of B are exactly
those which are infinite dimensional. The above reference claims this retraction
property holds for any infinite dimensional normed linear space. I think it is
fairly clear to see from the above example that this is not a surprising assertion.
Recall that one of the proofs of the Brouwer fixed point theorem used the non
existence of such a retraction, obtained using integration theory, to prove the
theorem.
We let K be a closed convex subset of X a Banach space and let
------
f be continuous, f : K → K,and f (K ) is compact.
Lemma 16.5.2For each r > 0 there exists a finite set of points
-----
{y1,⋅⋅⋅,yn} ⊆ f (K )
and continuous functions ψ_{i}defined on f
(K )
such that for x ∈f
(K )
,
∑n
ψi(x) = 1, (16.5.18)
i=1
(16.5.18)
ψi(x) = 0 if x ∕∈ B (yi,r),ψi(x) > 0 if x ∈ B (yi,r).
If
∑n
fr(x) ≡ yiψi(f (x)), (16.5.19)
i=1
(16.5.19)
then whenever x ∈ K,
∥f (x)− fr(x)∥ ≤ r.
Proof:Using the compactness of f
(K )
, there exists
{y ,⋅⋅⋅,y } ⊆ f (K-) ⊆ K
1 n
such that
{B (yi,r)}ni=1
covers f
(K )
. Let
+
ϕi(y) ≡ (r− ∥y − yi∥)
Thus ϕ_{i}
(y)
> 0 if y ∈ B
(yi,r)
and ϕ_{i}
(y)
= 0 if y
∕∈
B
(yi,r)
. For x ∈f
(K )
,
let
( n )− 1
ψ (x) ≡ ϕ (x)( ∑ ϕ (x)) .
i i j=1 j
Then 16.5.18 is satisfied. Indeed the denominator is not zero because x is in one of the
B
(yi,r)
. Thus it is obvious that the sum of these equals 1 on K. Now let f_{r} be given by
16.5.19 for x ∈ K. For such x,
is convex, and 16.5.18. It is just a convex combination of
things in B
(0,r)
. ■
Note that we could have had the y_{i}in f
(K )
in addition to being in f
(K)
.
This would make it possible to eliminate the assumption that K is closed later on. All
you really need is that K is convex.
We think of f_{r }as an approximation to f. In fact it is uniformly within r of f on K.
The next lemma shows that this f_{r} has a fixed point. This is the main result and comes
from the Brouwer fixed point theorem in ℝ^{n}. It is an approximate fixed point.
Lemma 16.5.3For each r > 0, there exists x_{r}∈ convex hull of f
(K )
⊆ K suchthat
fr(xr) = xr, ∥fr(x)− f (x)∥ < r for all x
Proof:If f_{r}
(xr)
= x_{r} and
∑n
xr = aiyi
i=1
for ∑_{i=1}^{n}a_{i} = 1 and the y_{i} described in the above lemma, we need
n n ( ( n )) n
fr(xr) = ∑ yiψi(f (xr)) = ∑ yjψj f ∑ aiyi = ∑ ajyj = xr.
i=1 j=1 i=1 j=1
Also, if this is satisfied, then we have the desired fixed point.
This will be satisfied if for each j = 1,
⋅⋅⋅
,n,
( ( n ))
a = ψ f ∑ ay ; (16.5.20)
j j i=1 ii
(16.5.20)
so, let
{ ∑n }
Σn −1 ≡ a ∈ ℝn : ai = 1,ai ≥ 0
i=1
and let h : Σ_{n−1}→ Σ_{n−1} be given by
( ( ) )
∑n
h(a)j ≡ ψj f aiyi .
i=1
Since h is a continuous function of a, the Brouwer fixed point theorem applies and there
exists a fixed point for h which is a solution to 16.5.20. ■
The following is the Schauder fixed point theorem.
Theorem 16.5.4Let K be a closed and convex subset of X, a normed linearspace. Let f : K → K be continuous and suppose f
(K )
is compact. Then f has afixed point.
Proof: Recall that f
(x )
r
−f_{r}
(x )
r
∈ B
(0,r)
and f_{r}
(x )
r
= x_{r} with x_{r}∈ convex hull
of f
(K)
⊆ K.
There is a subsequence, still denoted with subscript r such that f
(xr)
→ x ∈f
(K )
.
Note that the fact that K is convex is what makes f defined at x_{r}. x_{r}is inthe convex hull of f
(K )
⊆ K. This is where we use K convex. Then since f_{r} is
uniformly close to f, it follows that f_{r}
(xr)
= x_{r}→ x also. Thus x_{r} converges strongly to
x. Therefore,
f (x) = lim f (xr) = lim fr(xr) = lim xr = x. ■
r→0 r→0 r→0
We usually have in mind the mapping defined on a Banach space. However, the
completeness was never used. Thus the result holds in a normed linear space.
There is a nice corollary of this major theorem which is called the Schaefer fixed point
theorem or the Leray Schauderalterative principle [?].
Theorem 16.5.5Let f : X → X be a compact map. Then either
There is a fixed point for tf for all t ∈
[0,1]
or
For every r > 0, there exists a solution to x = tf
(x)
for some t ∈
(0,1)
suchthat
∥x∥
> r. (The solutions to x = tf
(x)
for t ∈
(0,1)
are unbounded.)
Proof: Suppose there is t_{0}∈
[0,1]
such that t_{0}f has no fixed point. Then t_{0}≠0.t_{0}f
obviously has a fixed point if t_{0} = 0. Thus t_{0}∈ (0,1]. Then let r_{M} be the radial
retraction onto B
(0,M )
. By Schauder’s theorem there exists x ∈B
(0,M )
such that
t_{0}r_{M}f
(x)
= x. Then if
∥f (x)∥
≤ M, r_{M} has no effect and so t_{0}f
(x)
= x which
is assumed not to take place. Hence
∥f (x )∥
> M and so
∥r f (x)∥
M
= M so
∥x∥
= t_{0}M. Also t_{0}r_{M}f
(x )
= t_{0}M
-f(x)-
∥f(x)∥
= x and so x =
ˆ
t
f
(x)
,
ˆ
t
= t_{0}
--M--
∥f(x)∥
< 1.
Since M is arbitrary, it follows that the solutions to x = tf
(x)
for t ∈
(0,1)
are
unbounded. It was just shown that there is a solution to x =
tˆ
f
(x)
,
ˆt
< 1 such that
∥x∥
= t_{0}M where M is arbitrary. Thus the second of the two alternatives holds.
■
Proof:Suppose that alternative 2 does not hold and yet alternative 1 also fails to
hold. Then there is M_{0} such that if you have any solution to x = tf
(x)
for t ∈
(0,1)
,
then
∥x∥
≤ M_{0}. Nevertheless, there is t ∈ (0,1] for which there is no fixed point for tf.
(obviously there is a fixed point for t = 0.) However, there is x such that for
M > M_{0},
x = t(r f (x)) = tM-f-(x-)= ˆtf (x), ˆt =-M-t < t
M ∥f (x)∥ ∥f (x)∥
We must have
∥f (x)∥
> M and r_{M}f
(x )
=
Mf-(x)
∥f(x)∥
since if
∥f (x)∥
≤ M, r_{M}f
(x )
= f
(x)
and there would be a fixed point for this t. Thus, letting M get larger and larger, there
are t_{M}∈
(0,1)
and x_{M},
∥x ∥
M
≤ M_{0} such that x_{M} = t_{M}f
(x )
M
,
∥f (x )∥
M
> M .
However, f is given to be a compact map so it takes B
(0,M )
0
to a compact set but this
shows that f must take this set to an unbounded set which is a contradiction. This
results from assuming there is t such that tf fails to have a fixed point for some t ∈
[0,1]
.
Thus alternative 1 must hold. ■
Next this is considered in the more general setting of locally convex topological vector
space. This is the Tychonoff fixed point theorem. In this theorem, X will be a locally
convex topological vector space in which every point is a closed set. Let ℬ be
the basis described earlier and let ℬ_{0} consist of all sets of ℬ which are of the
form B_{A}
(0,r)
where A is a finite subset of Ψ as described earlier. Note that for
U ∈ℬ_{0}, U = −U and U is convex. Also, if U ∈ℬ_{0}, there exists V ∈ℬ_{0} such
that
V + V ⊆ U
where
V +V ≡ {v1 + v2 : vi ∈ V}.
To see this, note
BA (0,r∕2)+ BA (0,r∕2) ⊆ BA (0,r).
We let K be a closed convex subset of X and let
------
f be continuous, f : K → K,and f (K ) is compact.
Lemma 16.5.6For each U ∈ℬ_{0}, there exists a finite set of points
-----
{y1⋅⋅⋅yn} ⊆ f (K)
and continuous functions ψ_{i}defined on f
(K )
such that for x ∈f
(K )
,
∑n
ψi(x) = 1, (16.5.21)
i=1
(16.5.21)
ψi(x ) = 0 if x ∕∈ yi + U,ψi(x) > 0 if x ∈ yi + U.
If
∑n
fU (x) ≡ yiψi(f (x)), (16.5.22)
i=1
(16.5.22)
then whenever x ∈ K,
f (x)− fU (x) ∈ U.
Proof:Let U = B_{A}
(0,r)
.Using the compactness of f
(K )
, there exists
{y ⋅⋅⋅y } ⊆ f-(K)
1 n
such that
{yi + U}ni=1
covers f
(K )
. Let
ϕi(y) ≡ (r− ρA(y − yi))+.
Thus ϕ_{i}
(y)
> 0 if y ∈ y_{i} + U and ϕ_{i}
(y)
= 0 if y
∕∈
y_{i} + U. For x ∈f
(K)
, let
( n )− 1
ψ (x) ≡ ϕ (x)( ∑ ϕ (x)) .
i i j=1 j
Then 16.5.21 is satisfied. Now let f_{U} be given by 16.5.22 for x ∈ K. For such
x,
∑
f (x)− fU (x) = (f (x)− yi)ψi(f (x))
{i:f(x)−yi∈U}