Sections of convex body translations, in a Banach space Let $K$ be a convex body in an infinite-dimensional real Banach space $E$, so $K$ is convex and has an interior.  Also suppose $K$ is closed and bounded.
Let $V$ be a finite-dimensional subspace of $E$ with volume measure $vol()$.
And suppose that $V$ intersects the interior of $K$, so $K \cap V$ has
an interior in $V$ and $vol(K \cap V) > 0$.
If $K$ is translated a small amount, I would like to know whether
that translation generates a small change in $K \cap V$.
Specifically for $x \in E$ define:
$$
f(x) ~:=~ vol( ~ (K\cap V) \vartriangle ~ ( (K+x)\cap V) ~)
~=~ vol( ~ (K \vartriangle ~ (K+x) )\cap V ~)
$$
where $\vartriangle$ is the symmetric difference of sets.
Obviously $f(0)=0$.


*

*Is $f(x)$ continuous at $0$ ?


I want the answer to be yes, but I suspect that it is no.
The vector $x$ can go in infinitely many independent directions,
so it may be possible to make a counter-example where
the curvature of the boundary of $K$ becomes unbounded, 
but I cannot see this clearly enough to make such an example.
Thank you for any answer.
 A: For brevity, I will write $\mu:={\rm vol}$. By assumption, there
is some $x_{0}\in K^{\circ}\cap V$, where $K^{\circ}$ denotes the
interior of $K$. Set $L:=K-x_{0}$ and note $0\in L^{\circ}\cap V$.
Furthermore, since the measure $\mu$ is invariant under translations
by elements of $V$, and since $x_{0}\in V$, we have for arbitrary
$x\in E$ that
\begin{align*}
 & \mu\left(\left(K\cap V\right)\Delta\left(\left(K+x\right)\cap V\right)\right)\\
 & =\mu\left(\left[\left(K\cap V\right)\Delta\left(\left(K+x\right)\cap V\right)\right]-x_{0}\right)\\
\left({\scriptstyle \text{since }V-x_{0}=V}\right) & =\mu\left(\left(L\cap V\right)\Delta\left(\left(L+x\right)\cap V\right)\right).
\end{align*}
Thus, we can (and will) assume from now on that $0\in K^{\circ}$.
Thus, there is some $\varepsilon>0$ with $B_{\varepsilon}\left(0\right)\subset K$.
Next, we make two important observations:


*

*Let $t\in\left(0,1\right)$ be arbitrary and $x\in E$ with $\left\Vert x\right\Vert <t\varepsilon$.
Then $\left\Vert \frac{1-t}{t}x\right\Vert <\varepsilon$ and thus
$\frac{1-t}{t}x\in B_{\varepsilon}\left(0\right)\subset K$. Next,
if $y\in K+x$, then $y-x\in K$ and thus
$$
\left(1-t\right)y=\left(1-t\right)\left(y-x\right)+t\cdot\frac{1-t}{t}x\in K.
$$
Thus, we have shown
$$
1_{K+x}\left(y\right)\leq1_{K}\left(\left(1-t\right)y\right)\qquad\forall y\in E\text{ and all }x\in E\text{ with }\left\Vert x\right\Vert <t\varepsilon.\qquad\left(\ast\right)
$$

*Let $t$ and $x$ as above and note $\left\Vert \frac{-x}{t}\right\Vert <\varepsilon$,
so that $\frac{-x}{t}\in B_{\varepsilon}\left(0\right)\subset K$.
Thus, if $\frac{y}{1-t}\in K$, then
$$
y-x=\left(1-t\right)\frac{y}{1-t}+t\cdot\frac{-x}{t}\in K.
$$
Thus, we have shown
$$
1_{K+x}\left(y\right)\geq1_{K}\left(\frac{y}{1-t}\right)\qquad\forall y\in E\text{ and all }x\in E\text{ with }\left\Vert x\right\Vert <t\varepsilon.\qquad\left(\dagger\right)
$$
Furthermore, it is not hard to verify (because of $0\in K$) that
$1_{K}\left(y\right)-1_{K}\left(\frac{y}{1-t}\right)\geq0$ and that
$1_{K}\left(\left(1-t\right)y\right)-1_{K}\left(y\right)\geq0$ for
all $y\in E$ and $t\in\left(0,1\right)$. All in all, if $\left\Vert x\right\Vert <t\varepsilon$,
then $\left(\ast\right)$ and $\left(\dagger\right)$ show
\begin{align*}
\left|1_{K+x}\left(y\right)-1_{K}\left(y\right)\right| & =\max\left\{ 1_{K+x}\left(y\right)-1_{K}\left(y\right),\,1_{K}\left(y\right)-1_{K+x}\left(y\right)\right\} \\
 & \leq\max\left\{ 1_{K}\left(\left(1-t\right)y\right)-1_{K}\left(y\right),\,1_{K}\left(y\right)-1_{K}\left(\frac{y}{1-t}\right)\right\} \\
\left({\scriptstyle \text{since everything is nonnegative}}\right) & \leq\left[1_{K}\left(\left(1-t\right)y\right)-1_{K}\left(y\right)\right]+\left[1_{K}\left(y\right)-1_{K}\left(\frac{y}{1-t}\right)\right]\\
 & =1_{K}\left(\left(1-t\right)y\right)-1_{K}\left(\frac{y}{1-t}\right)
\end{align*}
for all $y\in E$ and all $x\in E$ with $\left\Vert x\right\Vert <t\varepsilon$.
Using the change of variables formula in $V$, this finally implies
\begin{align*}
\mu\left(\left(K\cap V\right)\Delta\left(\left(K+x\right)\cap V\right)\right) & =\int_{V}\left|1_{K+x}\left(y\right)-1_{K}\left(y\right)\right|\,d\mu\left(y\right)\\
 & \leq\int_{V}1_{K}\left(\left(1-t\right)y\right)\,d\mu\left(y\right)-\int_{V}1_{K}\left(\frac{y}{1-t}\right)\,d\mu\left(y\right)\\
\left({\scriptstyle \text{with }n=\dim V}\right) & =\left(1-t\right)^{-n}\cdot\int_{V}1_{K}\left(z\right)\,d\mu\left(z\right)-\left(1-t\right)^{n}\cdot\int_{V}1_{K}\left(z\right)\,d\mu\left(z\right)\\
 & =\left[\left(1-t\right)^{-n}-\left(1-t\right)^{n}\right]\cdot\int_{V}1_{K}\left(z\right)\,d\mu\left(z\right),
\end{align*}
where $c:=\int_{V}1_{K}\left(z\right)\,d\mu\left(z\right)=\mu\left(K\cap V\right)$
is well-defined and finite, since $K$ is closed (and thus measurable)
and bounded. Since $\left(1-t\right)^{-n}-\left(1-t\right)^{n}\to0$
as $t\to0$, we are done.

Since this could be interesting for some people, here is my original answer. This was in response to the original question where it was only assumed that $K \cap V$ had nonempty interior (as a subset of $V$) and not that $K^\circ \cap V \neq \emptyset$.
This is false even in finite dimensions.
Consider $E = \Bbb{R}^2$, $V = \Bbb{R} \times \{0\}$ and $K = [-1,1] \times [-1,0]$, as well as $x_n = (0, - 1/n)$.
Then $K \cap V = [-1,1]$, but $K + x_n = [-1,0] \times [-1-\frac{1}{n}, -\frac{1}{n}]$, so that $(K + x_n) \cap V = \emptyset$ for all $n \in \Bbb{N}$.
Therefore, $f(x_n) = 2$, but $f(0) = 0$, even though $x_n \to 0$.
