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.
Best Answer
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
$$\|F(x_k,p_k)-F(x_0,p_0)\|\le \|F(x_k,p_k)-F(x_k,p_0)\|+\|F(x_k,p_0)-F(x_0,p_0)\|\le$$ $$\le \sup_{j\in\mathbb{N}}\|F(x_j,p_k)-F(x_j,p_0)\|+\|F(x_k,p_0)-F(x_0,p_0)\|=o(1).\qquad \square$$