I have been working on fixed point theory and was reading a paper by Thakur et. al., where they define Opial's condition. They say that a Banach space $E$ satisfies Opial's condition if for any sequence $\left( x_n \right)$ in $E$, which weakly converges to $x \in E$ we have
$$\lim\limits_{n \to \infty} \sup \| x_n - x \| < \lim\limits_{n \to \infty} \sup \| x_n - y \|,$$
for all $y \in E$ with $x \neq y$.
However, in the Opial's paper, where he gives this condition for the first time (as a theorem) in an Hilbert space, he mentions the following result:
For any sequence $\left( x_n \right)$ in $E$ which weakly converges to $x \in E$, we have
$$\lim\limits_{n \to \infty} \inf \| x_n - x \| < \lim\limits_{n \to \infty} \inf \| x_n - y \|,$$
for all $y \in E$ with $x \neq y$.
In the two definitions of the condition, all the terms are exactly the same, except Opial uses $\lim \inf$, while Thakur et. al. use $\lim \sup$.
Therefore, I tried to prove the equivalence of the two conditions, without which the further theorems in the paper by Thakur et. al. would have failed.
My attempt at proving the equivalence was as follows:-
We define two sequences $M_n = \inf \left\lbrace \| x_k - x \| : k \geq n \right\rbrace$ and $M_n' = \inf \left\lbrace \| x_k - y \| : k \geq n \right\rbrace$. Then, $\lim\limits_{n \to \infty} \inf \| x_n - x \| < \lim\limits_{n \to \infty} \inf \| x_n - y \|$ would mean $\lim\limits_{n \to \infty} M_n < \lim\limits_{n \to \infty} M_n'$. This is possible only if after some stage $n_0 \in \mathbb{N}$ we have
$$\| x_n - x \| < \| x_n - y \|, \forall n \geq n_0.$$
This would eventually mean that
$$\lim\limits_{n \to \infty} \sup \| x_n - x \| < \lim\limits_{n \to \infty} \sup \| x_n - y \|.$$
Similarly, we can prove the converse so that the equivalence is proved.
However, I highly doubt if this proof is correct. I especially doubt the second last step, where from $\lim\inf$ to inequality between individual elements. Also, even if that step is correct, I feel that after applying the $\lim\sup$, the strict inequality will not be preserved. Rather, it will be changed to $\leq$.
Comments on this attempt and possible corrections of the proof will be appreciated.
If $X$ is a Banach space that doesn't have Opial property ( by sup-definition) then there are $(x_n)_{n\in \mathbb{N}}$ and $x$ and $y\ne x$ such that $$x_n\to x \ \ \ (weakly) \ \ \ , \ \ \ limsup\lVert x_n-x\rVert \geq limsup\lVert x_n-y\rVert $$ There is a subsequent of $(x_n)_n$ like $(x_{f(n)})_n$ such that
$$ limsup\lVert x_n-x\rVert =lim\lVert x_{f(n)}-x\rVert $$
Therefore
$$ liminf\ \lVert x_{f(n)}-x\rVert= lim\ \lVert x_{f(n)}-x\rVert =limsup\lVert x_n-x\rVert\geq limsup\lVert x_n-y\rVert \geq limsup \lVert x_{f(n)}-y\rVert \geq liminf\ \lVert x_{f(n)}-y\rVert $$
So $X$ doesn't have Opial property (by inf-definition).
We can use the same arguments about the converse(If you want I can write it). So the definitions are equivalent.