Isoperimetric inequality for $\epsilon$-expansion of a set only along a certain subspace Let $\gamma_n$ be the standard gaussian distribution on $\mathbb R^n$. Let $V$ be a $k$-dimensional subspace of $\mathbb R^n$. Finally let $A$ be any (nonempty) Borel subset of $A$ with $\gamma_n(A) = p > 0$, and and for any $\epsilon\ge 0$, let $A^\epsilon_V$ denote the directional $\epsilon$-expansion of $A$ along $V$, defined by
$$
\begin{split}
A_V^\epsilon &:= A + V \cap B(x;\epsilon)\\
&= \{x \in \mathbb R^d \mid x = v + a,\text{ for some }a \in A,\text{ and }v \in V\text{ with }\|v\| \le \epsilon\}.
\end{split}
$$
Note that if $k = n$ (i.e if $V = \mathbb R^n$), then
$$
A^\epsilon_V = A + B(x;\epsilon) = \{x \in \mathbb R^n \mid d(x,A) \le \epsilon\},
$$
the usual $\epsilon$-expansion of $A$, and the isoperimetric inequality tells us that
$$
\gamma_n(A^\epsilon) \ge \Phi(\Phi^{-1}(p)+\epsilon).
\tag{1}
$$

Question 1. Is there an anologous bound for $\gamma_n(A_V^\epsilon)$ in terms of $p$, $k$, and $\epsilon$ ?

Now, let $V$ be drawn from the Haar measure on grassmanian of $k$-dimensional subspaces of $\mathbb R^n$.

Question 2. What is a good lower-bound (say, of the type (1)) of $\mathbb E_V[\gamma_n(A^\epsilon_V)]$ in terms of $p$, $k$, and $\epsilon$ ?

Note. I'd already be very happy for pointers papers studying these kinds of problems (directional expansions, etc.) or related things.
 A: $\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}\newcommand{\Ga}{\Gamma}$In Question 2 you wanted a good lower bound on $E_V P(X\in A_V^\ep)$, where $\ep>0$ and $X$ is a standard normal random vector in $\R^n$ and $E_V$ means the averaging over all $k$-dimensional subspaces $V$ of $\R^n$.
Consider first the case when the set $A$ one of the two half-spaces to one side of an affine hyperplane in $\R^n$. Then, in view of this answer, for each $x\in\R^n$,
\begin{equation*}
    E_V 1(x\in A_V^\ep)=Q_{n,k}\Big(\frac{d(x,A)^2}{\ep^2}\Big), \tag{1}
\end{equation*}
where $1-Q_{n,k}$ is the cdf of the beta distribution with parameters $k/2,(n-k)/2$ and $d(x,A)$ is the shortest distance from $x$ to $A$.
Equality (1) will hold if $A$ is bounded by a finite number of affine hyperplanes in $\R^n$ and the shortest distance from $x$ to the intersection of any two of these affine hyperplanes is $>\ep$.
So, for such $A$ and $\ep\downarrow0$,
\begin{equation*}
    E_V P(X\in A_V^\ep)=EE_V 1(X\in A_V^\ep)=EQ_{n,k}\Big(\frac{d(X,A)^2}{\ep^2}\Big)+O(\ep^2). 
\tag{1'}
\end{equation*}
Further, by the isoperimetric inequality,
\begin{equation*}
    P(0<d(X,A)\le\ep)\ge g(p)\ep\,(1+o(1)),
\end{equation*}
where $p:=P(X\in A)$ and
\begin{equation*}
    g(p):=\frac d{d\ep}\,\Phi(\Phi^{-1}(p)+\ep)\Big|_{\ep=0}=\vpi(\Phi^{-1}(p)), 
\end{equation*}
where $\vpi:=\Phi'$, the standard normal pdf.
So,
\begin{equation*}
\begin{aligned}
    &E_V P(X\in A_V^\ep)-P(X\in A) \\ 
&   =\int_0^\ep Q_{n,k}\Big(\frac{\de^2}{\ep^2}\Big)P(d(X,A)\in d\de) +O(\ep^2) \\  
&\ge\int_0^\ep Q_{n,k}\Big(\frac{\de^2}{\ep^2}\Big)g(p)\,d\de\,(1+o(1)) +O(\ep^2) \\  
&=c_{n,k}\,\ep g(p)\,(1+o(1)) +O(\ep^2),   
\end{aligned}
\end{equation*}
where
\begin{equation}
    c_{n,k}:=\int_0^1 Q_{n,k}\big(z^2\big)\,dz
    =\frac{\Ga((k+1)/2)/\Ga(k/2)}{\Ga((n+1)/2)/\Ga(n/2)}\Big[\sim\sqrt{\frac kn}\text{ if }n>k\to\infty\Big]. 
\end{equation}
Thus,
\begin{equation*}
    \liminf_{\ep\downarrow0}\frac{E_V P(X\in A_V^\ep)-P(X\in A)}\ep
    \ge c_{n,k}\,\vpi(\Phi^{-1}(p)). \tag{2}
\end{equation*}
Integrating this, we get
\begin{equation*}
    E_V P(X\in A_V^\ep)\ge\Phi\big(\Phi^{-1}(p)+c_{n,k}\,\ep\big)
\end{equation*}
for all real $\ep>0$.
Inequality (2) turns into the equality if the set $A$ is one of the two half-spaces to one side of an affine hyperplane in $\R^n$.
By approximation, (2) should hold at least for all convex $A$ and probably for all Borel $A$.

As for Question 1, the only lower bound on $P(X\in A_V^\ep)$ in general is the trivial lower bound $p=P(X\in A)$ -- for any natural $n\ge2$, any natural $k<n$, any real $\ep>0$, and any $p\in(0,1)$. Indeed, let the set $A$ be one of the two half-spaces to one side of an affine hyperplane in $\R^n$ such that $P(X\in A)=p$. Let $V$ be any $k$-dimensional subspace of $\R^n$ parallel to the mentioned affine hyperplane. Then $A_V^\ep=A$ for all real $\ep>0$, and hence $P(X\in A_V^\ep)=P(X\in A)$.
A: I provide a complete solution for the case where $A$ is the intersection of $N = \mathcal O(\mathrm{poly}(n))$ half-spaces $H_i := \{x \in \mathbb R^n \mid x^\top w_i \le b_i\}$, where each $w_i$ is a unit-vector in $\mathbb R^n$ and $b_i \in \mathbb R$.
At the moment, I'm unable to extend the argument to arbitrary convex sets.


Claim 1. For any subspace $V \subseteq \mathbb R^n$, define $r(V) := \min_{i \in [N]} \|P_Vw_i\| \in [0,1]$, where $P_V$ is the orthogonal projector for $V$.
Then, we have the set inclusion
$$
A^{r(V)\cdot\epsilon} \subseteq A^\epsilon_V,
$$
where $A^\delta := \{x \in \mathbb R^n \mid \mbox{dist}(x,A) \le \delta\}$ is the usual $\delta$-expansion of $A$ at level $\delta>0$.

Proof. If $\epsilon = 0$, both sets are equal to $A$ and there is nothing to prove. Otherwise, let $x \in A^{r(V)\epsilon}$. Then, $\mbox{dist}(x,A) \le r(V) \epsilon$. Now, it is well-known that the distance from a point to the intersection of a collection of sets is at least equal to the largest distance between the point and any of the sets. Using this fact, one computes
$$
\begin{split}
\frac{1}{\epsilon}\cdot\max_{i \in [N]}(x^\top w_i-b_i)_+ &= \frac{1}{\epsilon}\cdot \max_{i \in [N]} \mathrm{dist}(x,H_i)\\
&\le \frac{\mbox{dist}(x,A)}{\epsilon}\\
&\le r(V)\\
&:= \min_{i \in [N]}\|P_Vw_i\|\\
&= \min_{i \in [N]}\sup_{v \in V \cap B_d}v^\top w_i\\
&= \sup_{v \in V \cap B_d}\min_{i \in [N]}v^\top w_i.
\end{split}
$$
Thus, $\exists v \in V \cap B_d$ such that $x^\top w_i - b_i \le \epsilon v^\top w_i$, i.e $(x-\epsilon v)^\top w_i \le b_i$ for all $i \in [N]$. We conclude that $x \in A^\epsilon_V$. $\quad\quad\Box$
Let $p:=\mathbb P(X \in A)$. Since $X$ has gaussian concentration, we have
$$
\mathbb P(0 < \mbox{dist}(x,A) \le \epsilon) \ge g(p)\epsilon(1+o_\epsilon(1)),
$$
where $g(p):= \dfrac{{\rm d}}{{\rm}d\epsilon}\Phi(\Phi^{-1}(p)+\epsilon)\bigg|_{\epsilon=0} = \varphi(\Phi^{-1}(p))$ is the gaussian isoperimetric profile.

Strategy. With these auxiliary results in place, the rest of the proof is analogous to user @Iosif's post, starting from after formula (1'),

like so
$$
\begin{split}
\mathbb P(X \in A^\epsilon_V ) - P(X \in A) &\ge \mathbb P(X \in A^{r(V)\epsilon}) - P(X \in A)\\
&= \int_0^{r(V)\epsilon} \mathbb P(\mbox{dist}(X,A) \in {\rm d}\delta)\\
&\ge \int_0^{r(V)\epsilon} g(p){\rm d}\delta(1+o_\epsilon(1))\\
&= r(V)\epsilon g(p)(1+o_\epsilon(1)).
\end{split}
$$
We deduce that
$$
\liminf_{\epsilon \to 0^+}\frac{\mathbb P(X \in A^\epsilon_V ) - P(X \in A)}{\epsilon} \ge r(V)g(p) = r(V)\varphi(\Phi^{-1}(p)).
$$
Integrating w.r.t $\epsilon$ then gives the following

Result for general subspace. For any subspace $V \subseteq \mathbb R^n$, it holds that
$$
\mathbb P(X \in A^\epsilon_V) \ge \Phi(\Phi^{-1}(p)+r(V)\epsilon).
$$

The following corollary solves our initial problem.

(Result for random subspace). Let $V \subseteq \mathbb R^n$ be a uniformly-random $k$-dimensional subspace of $\mathbb R^n$. Then, for any $t \in (0,\sqrt{k/n})$ it holds w.p. $1-e^{-\Omega(t^2d)}$ that
$$
\mathbb P(X \in A^\epsilon_V) \ge \Phi(\Phi^{-1}(p)+(\sqrt{k/n}-t)\epsilon).
$$

Proof. For each $i \in [N]$, the quantity $\|P_V w_i\|^2 = w_i^\top P_V w_i$ has gaussian concentration around its expected value of $\mathbb E_V[\mbox{trace}(P_V)] = k/n$. It then follows from a basic union bound (since $N$ is only polynomial in $n$) that $r(V) \ge \sqrt{k/n}-t$ w.p $1-e^{-\Omega(t^2d)}$ over $V$. The claim then follows from the previous result. $\quad\quad\Box$.
A: This post solves the problem (hopefully) in case where $A$ is a closed convex set with "sufficiently smooth" boundary.

Preliminaries
Let $S_{n-1}$ be the unit-sphere in $\mathbb R^n$ and consider the mapping $u_A:\mathbb R^n \setminus A \to S_{n-1}$ defined by
$$
u_A(x):= (x-\Pi_A(x))/d_A(x),
$$
where $\Pi_A(x)$ is the closest point in $A$ from $x$ and $d_A(x):=\|x-\Pi_A(x)\| = \inf_{a \in A}\|x-a\|$ is the distance of $x$ from $A$. It is a classical result that $d_A$ is convex and continuously differentiable on the open set $\mathbb R^n \setminus A$, with derivative given by $\nabla d_A(x) = u_A(x)$. We will need the following condition.

Smoothness condition (SC1). $u_A$ is $L$-Lipschitz on $\mathbb R^n \setminus A$, for some $L \in [0,\infty)$.

As we will see later, the value of the smoothness condition (SC1) is that it allows us to compare $A^\varepsilon_V$ and $A^\varepsilon$ via differential calculs.
Examples and non-examples.
Thanks to this post, we know the following: https://mathoverflow.net/a/412746/78539

*

*(1) If $A$ is convex body (i.e compact and convex) of the form $A := \{x \in \mathbb R^n \mid f(x) \le 0\}$ for some $\mathcal C^2$ convex function with
$\min f < 0$, then the smoothness condition (SC1) is satisfied.

*(2) If $A$ is such that the polar cone of $A-x$ has dimension $\ge 2$ for some $x \in \partial A$ (e.g, if $A$ is a closed convex polytope with at least one vertex), then the smoothness condition (SC1) is not satisfied.


The result and proof
We now state the main theorem which will be proved in the remainder of this post

Theorem. Suppose $A$ is a closed convex set which satisfies the smoothness condition (SC1). Let $V$ be any subspace of $\mathbb R^n$ (random or not!) such that
$$
\mathbb P(\|P_Vu_A(x)\| \ge \alpha) \ge \beta,\text{ for all }x \in \mathbb R^n \setminus A,
\tag{*}
$$
where $P_V$ is the orthogonal projector for $V$ and $\alpha,\beta \in (0,1]$ are constants.
Then, for any $\varepsilon \in [0,2\alpha/L]$, we have the lower-bound
$$
\mathbb E_V \mathbb P_X(X \in A^\varepsilon_V) \ge \Phi(\Phi^{-1}(p)+\alpha\varepsilon)\cdot \beta,
$$
where $p:=\mathbb P(X \in A)$.

Of course, the above result can be nontrivial only when $\beta > p$.
Proof.
For any $x \in \mathbb R^n$ and sufficiently small $\varepsilon > 0$, we have the following chain of implications
$$
\begin{split}
x \not\in A^\epsilon_V &\implies x-\varepsilon v \not \in A\,\forall v \in V \cap B_n\\
&\implies d_A(x-\varepsilon v) > 0\,\forall v \in V \cap B_n\\
&\implies d_A(x) > \varepsilon v^\top u_A(x)-L\varepsilon\|v\|^2/2,\forall v \in V \cap B_n\\
&\implies d_A(x) > \varepsilon v^\top u_A(x)-L\varepsilon^2/2,\forall v \in V \cap B_n\\
&\implies d_A(x) > \varepsilon z^\top P_Vu_A(x) - L\varepsilon^2/2\,\forall z \in B_n\\
&\implies d_A(x) > \varepsilon \|P_Vu_A(x)\|-L\varepsilon^2/2.
\end{split}
$$
where

*

*the 3rd line is because the distance-to-$A$ function $d_A$ is differentiable on $\mathbb R^n \setminus A$, and its derivative $u_A$ is $L$-Lipschitz by hypothesis, and

*the fifth line is because $P_V z \in V \cap B_n$ for all $z \in B_n$.

By a contrapositive argument, we have established that
$$
\begin{split}
A^\varepsilon_V\setminus A &\supseteq \{x \in \mathbb R^n \setminus A \mid d_A(x) \le \varepsilon \|P_Vu_A(x)\|-L\varepsilon^2/2\}\\
& \supseteq \{x \in \mathbb R^n \mid 0 < d_A(X) \le \alpha\varepsilon-L\varepsilon^2/2,\, \|P_Vu_A(x)\| \ge \alpha\}.
\end{split}
\tag{1}
$$
Thus, for each $x \in \mathbb R^n$, sufficiently small $\varepsilon > 0$, we have
$$
\begin{split}
\mathbb E_V[1_{\{x \in A^\varepsilon_V\setminus A\}}] 
&\ge 1_{\{0 < d_A(x) \le \alpha \varepsilon - L\varepsilon^2/2\}}\cdot 1_{\{x \in \mathbb R^n\setminus A\}} \cdot \mathbb P_V(\|P_Vu_A(x)\| \ge \alpha)\\
&= 1_{\{0 < d_A(x) \le \alpha \varepsilon - L\varepsilon^2/2\}}\cdot \beta,
\end{split}
$$
Taking expectations w.r.t $X$ then gives
$$
\begin{split}
\mathbb E_V\mathbb P_X(X \in A^\varepsilon_V) - \mathbb P(X \in A) &= \mathbb E_V\mathbb P_X(X \in A^\varepsilon_V\setminus A)\\
&= \mathbb E_X\mathbb P_V(X \in A^\varepsilon_V\setminus A)\\
&= \mathbb E_X\mathbb E_V[1_{\{X \in A^\varepsilon_V\setminus A\}}]\\ &\ge \mathbb E_X[1_{\{0 < d_A(X) \le \alpha\varepsilon-L\varepsilon^2\}}]\cdot \beta\\
&=\mathbb P(0 < d_A(X) \le \alpha\varepsilon-L\varepsilon^2)\cdot\beta\\
&= \mathbb P(0 < d_A(X) \le \alpha\varepsilon-L\varepsilon^2)\cdot\beta.
\end{split}
$$
Now, for sufficiently small $\varepsilon$, one has
$$
\mathbb P(0  < d_A(X) \le \alpha\varepsilon-L\varepsilon^2/2) \ge \alpha g(p)\varepsilon(1+o_\varepsilon(1)),
$$
where $g$ is the gaussian isoperimetric profile defined in the other answers. The remainder of the proof follows exactly the same path as the others. $\Box$

Corollary. Let $V$ be uniform over the Grassmannian of $k$-dimensional subspaces of $\mathbb R^n$. Then, under the above smoothness condition, it holds for every $t \in (0,\sqrt{k/n})$ and $\varepsilon \in [0,2\alpha_t/L]$ with $\alpha_t := \sqrt{k/n}-t$ that
$$
\mathbb E_V\mathbb P_X(X \in A^\varepsilon_V) \ge \Phi(\Phi^{-1}(p)+\alpha_t\varepsilon)\cdot (1-e^{-\Omega(t^2d)}).
$$

Proof. Follows from previous theorem and the fact that
$$
\inf_{x \in \mathbb R^n\setminus A}\mathbb P(\|P_Vu_A(x)\| \ge \sqrt{k/n}-t) \ge \inf_{\theta \in S_{n-1}}\mathbb P(\|P_V\theta\| \ge \sqrt{k/n}-t) \ge 1-e^{-\Omega(t^2d)},
$$
where the last step is via chi-squared concentration (as in the other answers). $\quad\quad\quad\Box$
