Continuity with respect to Hausdorff metric, also with intersection? Let $H$ be a real Hilbert space and let $v\in H$, $\|v\|=1$. Now, consider the affine subspaces $A(t) := tv + v^\perp$ of codimension $1$, $t\in\mathbb R$. Obviously
$$
d_H(A(t),A(s)) = |t-s|
$$
for all $t,s\in\mathbb R$. Hence, $t\mapsto A(t)$ is Lipschitz-continuous with respect to the Hausdorff metric. That was quite easy.
However, what I really would like to show is that $f : t\mapsto A(t)\cap B$ is continuous, where $B$ is a bounded, closed, and convex subset of $H$ and $f$ is defined on $K := \{t\in\mathbb R : A(t)\cap B\neq\emptyset\}$, which is a closed set.
I've tried a lot, but I can't seem to prove this. So, I'd be happy if anyone had an idea. For clarification: the Hausdorff distance between two sets $A,B\subset H$ is defined by
$$
d_H(A,B) := \max\left\{\sup_{a\in A}\operatorname{dist}(a,B),\,\sup_{b\in B}\operatorname{dist}(b,A)\right\}.
$$
 A: The answer is almost yes. The function $f$ is continuous on the interior of the set of times such that it is non-empty.
Indeed, pick a reference point $x=(t_0 v+w)\in A(t_0)\cap B$, where $w\in v^{\perp}$, let $s<t_0$ and $t=\alpha s+(1-\alpha)t_0$ and observe $y=sv+w'\in f(s)$ where $w'\in v^{\perp}$. Then,
$$\alpha y+(1-\alpha)x=(\alpha s+(1-\alpha)t_0)v +\alpha w'+(1-\alpha)w,
$$
implying that $\alpha y+(1-\alpha)x\in A(t)$ and since $B$ is convex, we also get that $\alpha y+(1-\alpha)x\in f(t)$.
Hence, $$
d(y, f(t))\leq \|y-(\alpha y+(1-\alpha x))\|=(1-\alpha) \|y-x\|=(\frac{t-s}{t_0-s})\|x-y\|\leq \frac{t-s}{t_0-s} diam(B)
$$
Since this bound is uniform in $y$ we deduce that $\sup_{y\in f(s)}d(y,f(t))\leq \frac{t-s}{t_0-s} diam(B)$. Picking another reference point $z\in A(s_0)\cap B$ and running the argument backwards establishes
$$
d_H(f(t),f(s))\leq \max \left\{\frac{t-s}{t_0-s},\frac{t-s}{t-s_0} \right\} diam(B)
$$
for $t,s\in (s_0,t_0)$.
Unfortunately, continuity might fail at $t=\sup K$. Indeed, assume that $H$ is $\ell^2(\mathbb{N}_0)$ and let $v=v_0=(\delta^0_n)_{n\in \mathbb{N}_0}$ and $v_k:=(\delta^k_n)_{n\in \mathbb{N}_0}+(1-2^{-k})v_{0}$ for $k\geq 1$. Then, define $B$
$$
B:=\overline{conv} \{v_k|k\in \mathbb{N}_0\}
$$
Then, $B$ is closed, convex and bounded (it's contained in $B(0,3)$). We'll argue that $A(1)\cap B=\{v\}$. Having shown that, we'll just observe that $d(v_k, v)\geq 1$, proving that $d_H(A(1-2^{-k})\cap B,A(1) \cap B)\not \to 0$ as $t\to 1$.
So, indeed if $y\in A(1)\cap B$, then there exists $y_j=\sum_{k=0}^{N_j} \alpha_{k,j} v_k$ such that $\sum_{k=0}^{N_j}\alpha_{k,j}=1$, each $\alpha_{k,j}$ is positive and $y_j\to y$. Now, the $0$'th component of $y_j$ is at most
$$
(1-2^{-k_0})\alpha_{k_0,j}+\sum_{k\neq k_0} \alpha_{k,j}
$$
since the $0$'th component of $v_k$ is equal to $1-2^{-k}$. Since the $0$'th component of $y$ is equal to $1$, we get that $\alpha_{k,j}$ must go to $0$ for every $k\geq 1$.
Now, since $\alpha_{k,j}$ is equal to the $k$'th component of $y_j$ and the coordinate projections are continuous, we get that the $k$'th component of $y$ is $0$ for every $k\geq 1$. Thus, we get that $y=v$ and thus, that $A(1)\cap B=\{v\}$.
