To state our main result we need some definitions. Recall that a Banach
space X has the Uniform Opial Property [P3] if for every c>0
there exists r=r(c)>0 such that
for every
with
and every weakly null sequence
in X with 
.
To any Banach space X we can associate its modulus of Opial [LTX] by
where 
and the infimum is taken over all with
and all weakly null sequences
in X with .
In order to simplify the statement of the following theorem,
for 
we define the number
Since the well known inequality (see [X])
holds for all , then 
.
Proof
First, let us observe that we can suppose that there exists a>0 such that
Suppose, for a contradiction, that Y lacks the WFPP. Then there
exists a nonempty, convex and weakly compact set
and a fixed point
free 
-nonexpansive mapping
such that K is minimal for T. Since T has no
fixed point in K, then d=diam (K) >0 and we can suppose that d=1. Let
be an afps for T in K. Since every subsequence of is
again an afps for T, we can suppose that is an afps which is weakly
convergent and then, by translation of K, that K is minimal for T,
and that is weakly null.
Consider the subset 
of 
defined by
where 
.
It is easy to see that
is a nonempty, closed, convex and
-invariant set. Since
,
we have, by Lemma 1, that
Let 
be any element of . Since K is
weakly compact, there exists a subsequence 
of 
such that
and such that converges weakly to 
. By passing to subsequences,
we can assume that the following limits exist:
Fix 
. Since the sequence 
converges weakly to
, then
so that
Hence, we have that
Since 
is an afps for T, we have that
, by the Goebel-Karlovitz lemma, and then
>From this we get that
Since t<1, we can consider the sequence 
defined by
Then is weakly
convergent to 0 and 
. By the definition
of 
we have that
On the other hand, since
,
we have that
Hence, we have that
and then 
.
Pick any 
. Since
, there exists a subsequence 
of
such that
for every positive integer k.
On the other hand, we have that
and also that
>From the above facts we conclude that
and, since is arbitrary,
Finally, we get that
which contradicts (1).
Proof
Since 
is an euclidean norm, we have that
for all and then
.
On the other hand,
for all 
and then
By the Theorem, we only have to show that
i.e. that there exists such that
which is equivalent to
The left hand side F(a) in the above inequality takes its maximum value at
. Then a solution of the above inequation
is every B>1 such that
and this happens if 
.
In order to obtain a similar result for 
when 
and
we will need to find
the maximum a of a positive real function
defined for .
It is easy to see that
On the other hand, for nonnegative a
Since the term 
is positive,
the sign of the derivative F'(a) is the same
that the sign of the continuous function
We have
Moreover, for p>2
Therefore, there exists 
in the open interval 
, for
which 
. By arguments
from elementary Calculus, it is straightforward
to see that is the unique maximum of 
. One can search also
an interval for this maximum in the case 1<p<2, but we do not repeat
this for shake of brevity.
Proof
Since (see [X])
for all , then
.
On the other hand, (see [Do]), for all
By the Theorem,
we only have to show that there exists such that
i.e.
which is equivalent to
The left hand side 
in the above inequality takes
its maximum value
at some 
. Then a solution of the above inequation
is every B>1 such that
and this happens if 
.
The effective computation of the constant 
in the above
corollary seems to be hard, spcecially if p is not a natural number.
Nevertheles, it is clear that
The right hand side in this inequality
fournishes bounds of stability in which are greater than
, at least for
p<6.
Let X be a Banach space with Opial modulus .
In [LTX] the authors proved that
is continuous on 
and that, for 
,
Then, if satisfies the condition 
, for 
, we have
Therefore, for the Banach spaces X with , 
, ( ).
It is obvious that, for such spaces, the stability bound given by our theorem is M(X).
Spaces verifying are, for example,
(see [LTX], Remark 3.1.), and
the Bynum spaces 
( ).
As a counterpart of the above remark we can state the following result.
Proof
Since is a continuous function, there exists 
such that
. By definition we obtain
Now, we can apply the theorem for B=1. Since for every a> 0
one has 
, we have
Notice that in the Theorem 3.3. of [Do] has been
proved that 
implies M(X)>1, for reflexive Banach space,
in some sense, a dual version of last Corollary.
Acknowledgement. The authors would like to thank Professor T. Domínguez Benavides for his comments on the first version of this paper, mainly for the result of Corollary 3.