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Let $(X,d)$ be a complete metric space, $a \in X$ and $r \gt 0$. Let $f: B(a,r) \rightarrow X$ be a contraction with a contraction constant $L$, $d(f(x),f(y)) \lt Ld(x,y)$ , for all $x, y \in B(a,r)$. Assume moreover that $d(f(a),a) \lt (1-L)r$.

The aim is to show that the function $f$ has a unique fixed point in $B(a,r)$. In order to apply the fixed point theorem we need to restrict $f$ to a subset of $B(a,r)$ where the values of $f$ on this subset remain in it then we can apply the theorem on this subset. The suggestion in the question was to prove that there is some $r_0\in ]0,r[$ such that $d(f(a),a)) \leq (1-L) r_0$. I cannot see how can I find a subset of this open ball where such an inequality will resemble the given one. Any hints or suggestions? Thanks!

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$d(f(a),a) <(1-L)r$ and so there eixts $r_0 <r$ such that $d(f(a),a) <(1-L)r_0$. (This is just continuity of the right hand side as a function of $r$).

Now $d(x,a) \leq r_0$ implies $$d(f(x),a)\leq d(f(x),f(a))+d(f(a),a) \leq Ld(x,a) +(1-L) r_0 \leq Lr_0 +(1-L)r_0=r_0$$ so $f$ maps the closed ball with center $a$ and radius $r_0$ into itself. Now apply fixed point theorem to the restriction of $f$ to this closed ball.

Kavi Rama Murthy
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  • I think I am missing the first point. I know how to proceed further but I don't understand how did we use continuity to reach this inequality: $d(f(a),a) \leq (1-L)r_0$? –  Aug 05 '21 at 03:58
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    @user7857462 If a real number $x$ satisfies $x <r$ then $x<r_0$ for some $r_0 <r$. As simple as that! Take $x$ to be $\frac {d(f(a),a)} {1-L}$. – Kavi Rama Murthy Aug 05 '21 at 05:03