Is the closed, bounded and convex subset version of Shauder-Tychonoff Fixed Point Theorem really in the literature?

In 1930, J. Schauder extended Brouwer's work to arbitrary Banach spaces by stating the theorem;

Schauder Fixed Point Theorem: Let $$K\subset E$$ be a compact convex set where $$E$$ is Banach and $$T :\,K\longrightarrow K$$ a continuous map. Then, $$T$$ has a fixed point.

However, there is another fixed point theorem called Shauder-Tychonov Fixed Point Theorem, it states like this;

Shauder-Tychonoff Fixed Point Theorem: Let $$E$$ be a Banach space and $$K\subset E$$ be a non-empty closed, bounded and convex set. Suppose that $$T:K\longrightarrow K$$ is completely continuous, then there exists $$x^*\in K\;\text{such that} \;Tx^*=x^*.$$

I have tried all my best to get this paper. I don't know which of the authors actually wrote the theorem. Several papers I found are misleading.

Question: Can anyone please, direct me to a link on the exact paper where Shauder-Tychonoff Fixed Point Theorem was derived? I need it for my literature review. Thanks.

• I guess you know this reference: Tychonoff, A. Ein Fixpunktsatz. (German) Math. Ann. 111 (1935), no. 1, 767–776. eudml.org/doc/159810 . There, he generalizes the Schauder theorem to locally convex spaces, which comes close to your desired theorem. According to mathscinet, it is the only publication of Tychonoff that has 'fix' in its title.
– daw
May 7, 2019 at 10:14
• @daw: I do know that reference. Since it's not written in German, I might not be able to interpret it. That brings me to the next question. How does local convexity of the space imply that $K$ is closed, bounded, convex and nonempty? I asked because you said it is the only one close to it. I, somehow, agree though! May 7, 2019 at 10:37
• The theorem is wrt to the same setting as Schauder: $K$ compact, convex, $T$ continuous, but $E$ lc space
– daw
May 7, 2019 at 11:14
• @daw: Sorry, I meant to say, "Since it's not written in English..." May 7, 2019 at 11:24
• @daw: So, that means it's wrong to put the above theorem in a paper, right? I have, however, found it in several journals. Are they also wrong? May 7, 2019 at 11:27

The desired theorem is not true in this full generality. Here is a counterexample:

The right-shift operator $$R$$ on $$l^1(\mathbb N)$$ is completely continuous (due to the Schur property of $$l^1$$). Now define $$T$$ as $$Tx = ( 1- \|x\|_{l^1} , Rx).$$ Then $$T$$ is completely continuous, maps elements from the closed unit ball to the closed unit ball, but has no fixed point: The image of $$T$$ is the boundary of the closed unit ball. If $$x$$ would be a fixed point of $$T$$, then by the definition of $$R$$, $$x$$ is a constant sequence. But the boundary of the unit ball does not contain a constant sequence.

• Kinldy check this paper out SIDNEY A. MORRIS, E. S. NOUSSAIR. The following theorem is in page $1;$ Strong version of Schauder's Theorem: Let $A$ be a closed convex subset of a Banach space and, $f$, a continuous map of $A$ into a compact subset of $A$. Then $f$ has a fixed point. Can I call this the desire theorem? May 7, 2019 at 10:43
• No: there the set is compact and $f$ continuous, while in your statement in the question the set is weakly compact and $f$ is completely continuous.
– daw
May 7, 2019 at 15:34