I have been reading Felix E. Browder's [Convergence Theorems for Sequence of Nonlinear Operators in Banach Space](https://link.springer.com/article/10.1007/BF01109805) and I was hoping I could find answers to a couple of questions I have about the paper. Consider Lemma 6.



> 
> Let $F$ be a closed convex subspace of a Hilbert space H. Let
> $\{u\_n\}$ be a sequence in $H$ such that
> 
> 
> 1. For each $f\in F$, $\|u\_n-f\|$ is a non-increasing sequence
> 2. Each weak limit point of the sequence $\{u\_n\}$ lies in $F$.
> 
> 
> Then $u\_n \to f\_0$ weakly for some point $f\_0$ in $F$.
> 
> 
> 


I can follow most of the proof. However, is the bound $\|u\_n\|\leq \|u\_1\|+\|f\|$ for a fixed $f$ necessarily correct? It seems that it should be $\|u\_n\|\leq \|u\_1\|+2\|f\|$. In any case, I am not sure how one obtains $p(f):=\lim\_{n\to \infty} \|u\_n-f\| \ge \|f\|-M\_0$ (I don't think he mentions what $M\_0$ stands for, but I strongly suspect it is the bound of $\|u\_{n}\|$, for which the inequality would make sense).


Also, how does lower semi-continuous $p$ assumes minimum at some point $f\_0$ of $F$? Is there is a version of the following theorem ([link](https://math.stackexchange.com/questions/3082099/show-that-lower-semi-continuous-function-attains-its-minimum-proof-verificati)) for closed convex subsets of Hilbert spaces and lower semi-continuous $p$?


Further, consider Lemma 7.



> 
> Let $X$ be a strictly convex Banach space. $U$ is a non-expansive mapping of a convex subset $C$ of $X$ into $X$. Then the fixed point set of $U$ in $C$ is convex.
> 
> 
> 


The proof starts off by considering two fixed points $f\_0$ and $f\_1$ of $U$ and trying to prove that $y\_t= (1-t)f\_0+tf\_1$ is a fixed point for $0\leq t \leq 1$. I have been able to follow the proof except that I don't see how the remark that $U(y\_t)$ lies on the line segment between $f\_0$ and $f\_1$ follows. Thanks!



On some convergence theorems by Felix E. Browder (1967)
**Concerning Lemma 6:** Clearly, this lemma is false if $F=\emptyset$ and, say, $u\_n=nu$ for some nonzero $u\in H$. Assume therefore that $F\ne\emptyset$.


The inequality $\|u\_n\|\le\|u\_1\|+\|f\|$ will not always hold in this setting -- consider e.g. the case when $u\_1=0$ and $u\_n=2f\ne0$ for $n\ge2$.


However, since $F\ne\emptyset$, there is some $f\_\*\in F$. Then $\|f\_\*-u\_n\|\le\|f\_\*-u\_1\|$ for all $n$, and hence $\|u\_n\|\le M:=\|f\_\*\|+\|f\_\*-u\_1\|$ and $\|f-u\_n\|\ge\|f\|-\|u\_n\|\ge\|f\|-M$.


So, $p(f)\ge\|f\|-M$, and hence $p(f)>p(0)$ if $\|f\|>r:=M+p(0)$. So, $$\inf\_F p=\inf\_{F\cap B\_r} p,$$
where $B\_r$ is
the closed ball of radius $r$ in $H$ centered at $0$. The ball $B\_r$ in the Hilbert space $H$ is [weakly compact](https://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem). The set $F$ is convex and closed and [hence weakly closed](https://math.stackexchange.com/questions/1151148/a-convex-subset-of-a-banach-space-is-closed-if-and-only-if-it-is-weakly-closed). So, $F\cap B\_r$ is weakly compact. Also, the lower semicontinuous **convex** function $p$ is [weakly lower semicontinuous](https://link.springer.com/content/pdf/10.1007%2F978-1-4419-9467-7_9.pdf). Thus, $p$ does attain a [minimum](https://en.wikipedia.org/wiki/Semi-continuity#Properties) on $F\cap B\_r$ and hence on $F$ (provided that $F\ne\emptyset$).




---


**Concerning Lemma 7:** The desired conclusion follows immediately from [third listed property equivalent to the strict convexity of the normed space](https://en.wikipedia.org/wiki/Strictly_convex_space).


