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Theorem.$\ $ Let $E$, $F$ be two Banach spaces, $U$ an open ball in $E$, $V$ an open ball in $F$ of center $y_0$ and radius $\beta$, and $v$ a continuous mapping of $U \times V$ into $F$. Suppose that there exists a constant $k$ with $0 < k < 1$ such that $\left|\left| v(x,y_0) - y_0\right|\right| < \beta(1-k)$ for all $x \in U$ and $$\left|\left|v(x,y_1) - v(x,y_2)\right|\right| \le k\left|\left|y_1 - y_2\right|\right|$$ for all $x \in U$ and all $y_1,y_2 \in V$. Then there exists a unique continuous mapping $f$ of $U$ into $V$ such that $$f(x) = v(x,f(x))$$ for all $x \in U$.

I'm trying to prove this using the

Banach fixed point theorem.$\ $ Let $F$ be a Banach space, $V$ an open ball in $F$ of center $y_0$ and radius $\beta$, and $v$ a mapping of $V$ into $F$. Suppose that there exists a constant $k$ with $0 < k < 1$ such that $\left|\left|v(y_0) - y_0\right|\right| < \beta(1-k)$ and $$\left|\left| v(y_1) - v(y_2) \right|\right| \le k\left|\left|y_1 - y_2\right|\right|$$ for all $y_1,y_2 \in V$. Then $v$ has a unique fixed point.

Here is what I have so far.

Proof.$\ $ Let $x \in U$. The mapping $v_x$ of $V$ into $F$ defined by $v_x(y) = v(x,y)$ satisfies the conditions of the Banach fixed point theorem. Let $f(x)$ denote its unique fixed point. This defines a mapping $f$ of $U$ into $V$ such that $f(x) = v(x,f(x))$ for all $x \in U$.

Question.$\ $ How do I show that $f$ is continuous?

Uniqueness is immediate.

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  • $\begingroup$ There is a general result that fixed points of contractions depend continuously on parameters, if the mappings depend continuously on them. This is an application of this result. $\endgroup$ – Hans Engler Jul 7 '16 at 23:30
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Let $(x_n)$ be a sequence which converges towards $x$, $\|f(x_n)-f(x)\|=\|v(x_n,f(x_n))-v(x,f(x))\|\leq \|v(x_n,f(x_n))-v(x_n,f(x))\|+\|v(x_n,f(x))-v(x,f(x))\|$.

$\|v(x_n,f(x_n))-v(x_n,f(x))\|\leq k\|f(x_n)-f(x)\|$. This implies that $\|f(x_n)-f(x)\|\leq \|v(x_n,f(x))-v(x,f(x))\|+ k\|f(x_n)-f(x)\|$.

We deduce that $(1-k)\|f(x_n)-f(x)\|\leq \|v(x_n,f(x))-v(x,f(x))\|$

We have $lim_n\|v(x_n,f(x))-v(x,f(x))\|=0$ since $v$ is continue. This implies that $lim_nf(x_n)=f(x)$.

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