Proving if $f_n(a)\to f(a)$ then $f(a)=a$ in $C_{[0,1]}$

Question

Exercise:For each positive integer $n$ let $f_n$ be a continuous function of [0,1] into itself $a\in[0,1]$ be such that $f_n(a)=a$ for all $n$. Further let $f$ be a continuous function of [0,1] into itself.

If $f_n\to f$ in $(C_{[0,1]},d)$ where d is the supremum metric. Prove that $a$ is a fixed point of $f$.

$\exists N\in\mathbb{N}$ such that $d(f_N,f)=\sup|f_N-f|<\epsilon$ for $\epsilon>0$.

As the function converges uniformly then $\sup|f_N(a)-f(a)|<\epsilon$.

$f_n$ is continuous for all $n$ so:

$f_N(B(a,\delta))\subseteq B(f_N(a),\epsilon)$

But as $\sup|f_N(a)-f(a)|<\epsilon$ so $f(a)\in B(f_N(a),\epsilon)$

So I think I can conclude $f(a)=a$

Questions:

Is my proof right? If not. What are the alternatives? Is my latest conclusion right? (I am not comfortable with that step).

Thanks in advance!

Answer 1

Suppose that $a$ were not a fixed point of $f$, then set $\varepsilon = |f(a)-a| >0$. We then have for this concrete $\varepsilon$ some $N$ such that for all $n \ge N$ we have $d(f_n,f) < \varepsilon$.

Clearly $|a - f(a)| = |f_n(a) - f(a)| \le d(f_n,f)$ by definition, but taking e.g. $n=N$ we get $\varepsilon = |a - f(a)| \le d(f_N,f) < \varepsilon$, which is a contradiction.

So $f(a) =a$ must hold.