Is the metric projection on a strictly convex space continuous with respect to the set? Let $X$ be a reflexive strictly convex Banach space and $C \subset X$ be a nonempty closed convex subset. Then the metric projection $P_X : X \rightarrow C$ is well-defined: $P_C(x)$ is the element satisfying
$$\|x - P_C(x)\| = \inf_{c \in C} \|x - c\|.$$
That is, $P_C(x)$ is the element of $C$ which best approximates $x$.
Now suppose we have a family $(C_t)_{t}$ of nonempty closed convex subsets of $X$. Is there some notion of distance between the sets $C_t$ such that for any $x \in X$ the function $f(t) = P_{C_t}(x)$ is continuous? That is, is there any notion of distance between convex sets which ensures that if two convex sets are close then the metric projections with respect to a given point must be close? What if we suppose that $X$ is uniformly convex?
 A: Let $X$ be a uniformly convex space and $x\in X$.Let me start by showing the continuity with respect to bounded, closed and convex sets.
We denote by $h$ the Hausdorff distance on the space $\mathcal{CB}(X)$ of bounded closed and convex subsets of $X$ and recall that for all sets $A,B\in\mathcal{CB}(X)$ the inequality
$$
    |d(x,A)-d(x,B)| \leq h(A,B)
$$
holds. Given $A\in\mathcal{CB}(X)$ and $\varepsilon>0$, we set
$$
    t = \frac{\varepsilon}{2(d(x,A)+1)} \qquad \text{and} \qquad \tilde{\delta} = \min\left\{\frac{\delta_X(t) d(x,A)}{2+\delta_X(t)}, \frac{\varepsilon}{2},1\right\}
$$
and note that $\tilde{\delta}>0$ since $X$ is uniformly convex. For $B\in\mathcal{CB}{X}$ with $h(A,B)<\tilde{\delta}$ we note that we may find $z\in B$ with $\|P_Ax-z\|<\tilde{\delta}$ since $d(P_Ax,B)\leq h(A,B)<\tilde{\delta}$. We now have
$$
 \|x-P_Bx\| = d(x,B) \leq d(x,A)+\tilde{\delta}
$$
and
$$
\|x-z\| \leq \|x-P_Ax\| + \|P_Ax-z\| \leq d(x,A)+\tilde{\delta}.
$$
Assume for a contradiction that
$$
 \|P_Bx-z\|\geq \frac{\varepsilon}{2} = (d(x,A)+\tilde{\delta}) \tilde{t} \qquad \text{with} \qquad \tilde{t} = \frac{\varepsilon}{2(d(x,A)+\tilde{\delta})} \geq t
$$
and observe that
\begin{align*}
    \left\|x-\frac{P_Bx+z}{2}\right\| &\leq (d(x,A)+\tilde{\delta}) (1-\delta_X(\tilde{t})) \leq (d(x,A)+\tilde{\delta}) (1-\delta_X(t)) \\
                                      & \leq d(x,A)-\tilde{\delta} + 2\frac{\delta_X(t) d(x,A)}{2+\delta_X(t)} - \delta_X(t) d(x,A)\\
                                      & < d(x,A)-\tilde{\delta}
\end{align*}
which, since $B$ is convex, contradicts $d(x,B) \geq d(x,A)- h(A,B) > d(x,A)-\tilde{\delta}$. In other we words, we have now shown that
$$
    \|P_Ax-P_Bx\| \leq \|P_Ax-z\|+\|z-P_Bx\| < \varepsilon,
$$
as required.
For unbounded closed and convex sets, the results then follows, if we assume that there is some closed ball $\bar{B}(x,r)$ with e.g. $r=2d(x,A)$ with the property that the intersection with the closed and convex sets is nonempty and that the intersections convergence in the Hausdorff distance.
