Let $X,Y$ be Hilbert spaces and $P$ a topological space$^1$ and $p\_0\in P$.


Let $f:X\times P\to Y$ be a *continuous* map such that
for any parameter $p\in P$, $f\_p:= f|\_{X\times \{p\}}:X\to Y$ is *smooth*.
Suppose also that $f\_p\to f\_{p\_0}$ in $C^3\_{loc}(X,Y)$ for $p\to p\_0$ (i.e. once we fix an arbitrary compact $K\subset X$, we have uniform convergence in $C^3$-norm over it)


Suppose that $Df\_{p\_0}(x\_0):X\to Y$ is an isomorphism, we would like to prove that under these assumptions there exists a continuous implicit function, i.e.



> 
> Show that exists $U\_{x\_0}\times U\_{p\_0}\subset X\times P$ product neighbourhood of $(x\_0,p\_0)$ and *continuous* function $x:U\_{p\_0}\to U\_{x\_0}$ such that the zero locus of $f|\_{U\_{x\_0}\times U\_{p\_0}}$ is the graph of $x$, $\{(x(p),p)\ | \ p\in U\_{p\_0}\}$.
> If false please provide an example
> 
> 
> 


Notice that the existence of the neighbourhood is already a difficult part. Indeed $C^1\_{loc}$ convergence implies that there is $U\_{p\_0}$ such that for any $p\in U\_{p\_0}$, $Df\_{p}(x\_0)$ is isomorphism so we can apply the inverse function theorem but the size of the "inversion"-neighbourhood can shrink to zero as $p\to p\_0$.
In order to prevent this, I thought to use the $C^3\_{loc}$ convergence, because we can estimate the radius of the neighbourhood using the second derivative in a neighbourhood of $x\_0$, however the convergence is only on compact sets therefore I do not know how to use it to produce a bound for $D^2f\_p(x)$ for $\|x-x\_0\|<r$.


For the continuity I hope that somebody knows a version of the inverse function theorem or contraction principle that works in this non-smooth setting, unfortunately $C^r\_{loc}(X,Y)$ is not even a metric space for $X,Y$ infinite dimensional.


$^1$ I am interested in the case where $P$ is a finite dimensional manifold so feel free to assume so if you can say something about this situation.



Implicit function theorem with continuous dependence on parameter
Assuming $P$ first countable, the standard contraction principle and elementary bounds are sufficient to conclude. You do not need higher regularity:


*Let $X$, $Y$ be Banach spaces, $P$ a topological space, $f:X\times P\to Y$. Assume that*


*i. $f:X\times P\to Y$ is continuous;*


*ii. $f(\cdot,p):X\to Y$ is differentiable, for any $p\in\ P$.*


*iii. $f(x\_0,p\_0)=0$ and $D\_1f(x\_0,p\_0)$ is invertible.*


*iv. $D\_1 f :X\times P \to L(X,Y)$ is continuous *in the pair* at $(x\_0,p\_0)$.*


**Proof:** Let $L:=[D\_1f(x\_0,p\_0)]^{-1}\in L(Y,X)$, and define $g:X\times P\to X$ by $g(x,p):=x-Lf(x,p)$, so that $f(x,p)=0$ iff $x=g(x,p)$. Since $D\_1g(x\_0,p\_0)=0$, and (by iv) $D\_1g$ is continuous at $(x\_0,p\_0)$, there are $r>0$ and a nbd $U$ of $p\_0$ in $P$ such that $\|D\_1g(x,p)\|\le 1/2$ for all $x\in \overline B(x\_0,r)$ and $p\in U$, which implies that for all $p\in U$ the maps $g(\cdot,p)$ are $1/2$ Lipschitz. Since $g(x\_0,p\_0)=x\_0$ and $g$ is continuous, choosing a smaller $U$ we may also assume $\|g(x\_0,p) -x\_0\|\le r/2$. Then for all $x\in \overline B(x\_0,r)$ and $p\in U$ we have
$$\|g(x,p)-x\_0\|\le \|g(x,p) -g(x\_0,p)\|+\|g(x\_0,p) -x\_0\|\le \frac12\|x-x\_0\|+\frac r2\le r.$$
So for all $p\in U$, the closed ball $\overline B(x\_0,r)$ is invariant for the contraction $g(\cdot,p)$, which has therefore a unique fixed point $x=\xi(p)$ in $\overline B(x\_0,r)$. In other words, for any $(x,p)\in \overline B(x\_0,r)\times U$ one has $f(x,p)=0$ iff $x=\xi(p)$. To show that $\xi:U\to \overline B(x\_0,r)$ is continuous at any $p\in U$, write for $q\to p$ in $U$
$$\|\xi(p)-\xi(q)\|=\|g(\xi(p),p)-g(\xi(q),q)\|\le \|g(\xi(p),p)-g(\xi(p),q)\|+\|g(\xi(p),q)-g(\xi(q),q)\|\le$$
$$\le \|g(\xi(p),p)-g(\xi(p),q)\|+\frac12\|\xi(p) -\xi(q) \|,$$
so $\|\xi(p)-\xi(q)\|\le 2\|g(\xi(p),p)-g(\xi(p),q)\|=o(1)$ as $q\to p.$ $\qquad\square$


$$ \* $$


Note that the assumption *iv* on the map $F:=D\_1g$ follows from your hypotheses, provided $P$ is assumed first countable:


*Let $X,P$ be first countable spaces, $(E,\|\cdot\|)$ a normed space, $F:X\times P\to E$. Assume that $F(\cdot,p\_0):X\to E$ is continuous at $x\_0$ and $P\ni p\mapsto F(\cdot,p)$ is continuous at $p\_0$ w.r.to the uniform convergence on compact sets (that is, $\|F(\cdot,p)-F(\cdot,p\_0)\|\_{\infty,K}=o(1)$ as $p\to p\_0$). Then $F$ is continuous at $(x\_0,p\_0).$*


Indeed, for any sequence $(x\_k,p\_k)\to (x\_0,p\_0)$ in $X\times P$ we have


$$\|F(x\_k,p\_k)-F(x\_0,p\_0)\|\le \|F(x\_k,p\_k)-F(x\_k,p\_0)\|+\|F(x\_k,p\_0)-F(x\_0,p\_0)\|\le$$
$$\le \sup\_{j\in\mathbb{N}}\|F(x\_j,p\_k)-F(x\_j,p\_0)\|+\|F(x\_k,p\_0)-F(x\_0,p\_0)\|=o(1).\qquad \square$$


