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The question below is related to the classical Browder-Goehde-Kirk fixed point theorem.

Let $K$ be the closed unit ball of $\ell^{2}$, and let $T:K\rightarrow K$ be a mapping such that $$\Vert Tx-Ty\Vert _{\ell^{4}}\leq\Vert x-y\Vert _{\ell^{3}}$$ for all $x,y\in K$.

Is it true that $T$ has fixed points?

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    $\begingroup$ It would be easier to read if you just wrote $\|Tx−Ty\|_4 \le \|x−y\|_3$. It took me a minute to get the point of your question since I could barely see the difference between the supersubscripts $\endgroup$
    – Mark Meckes
    Commented Mar 15, 2010 at 13:36
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    $\begingroup$ I don't think it does. I just meant to make sure other readers could easily tell what you were asking; I didn't downvote and I hope I didn't encourage anyone else to. $\endgroup$
    – Mark Meckes
    Commented Mar 15, 2010 at 13:48
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    $\begingroup$ @Fabrizio Polo I'm not claiming that. Just that, e.g., $FP(2,4,3)$ holds, via BGK. Please read carefully my comment. $\endgroup$
    – Ady
    Commented Apr 15, 2010 at 17:17
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    $\begingroup$ Would someone please state the classical Browder-Goehde-Kirk fixed point theorem? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jun 21, 2013 at 0:39
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    $\begingroup$ $FP(p,p,\infty)$ is false for any $p$. We can let $(Tx)_i = x_{i-1}$ for $i<1$ and $(Tx)_0 =(1/2)( 1 - \sum_i x_i^2)$. Then $||Tx||_{\ell_2}^2 = (1/4)(1-||x||_{\ell_2}^2)^2+ x_{\ell_2}^2 \leq 1$ since $x_{\ell_2}\leq 1$ and, because each coordinate is a $1$-Lipschitz function in the $\ell_2$ norm, the whole thing is a $1$-Lipschitz transformation from $\ell^\infty$ to $\ell^2$. $\endgroup$
    – Will Sawin
    Commented Dec 18, 2016 at 14:03

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