$\newcommand{\kk}{\mathbb{k}}
\newcommand{\mmm}{\mathfrak{m}}
\newcommand{\ppp}{\mathfrak{p}}
\newcommand{\qqq}{\mathfrak{q}}
\DeclareMathOperator{\gr}{gr}$Assume $G := \gr_\mmm(R)$ satisfies the following two conditions:

- $G$ has
*pure* dimension $d$.
- The zero ideal of $G$ has no embedded prime which does not contain the maximal ideal $G_+ := \oplus_{i \geq 1}\mmm^i/\mmm^{i+1}$.

**Theorem.** If $G$ satisfies the above conditions, then the answer is negative, i.e. if $f \in R$ is such that $\dim(\gr_\mmm(R)/\langle f^* \rangle) = d-1$, then $f$ is superficial.

Before proving the theorem, some examples (please doublecheck that the claims in the examples are correct - I have not checked these carefully):

- An example where Condition 1 fails: If $R$ is the localization at the origin of $\kk[x,y,z]/\langle xy, xz \rangle$, where $\kk$ is a field, then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*]/\langle x^*y^*, x^*z^* \rangle$. In particular, $d := \dim(R) = 2$, $\dim(\gr_\mmm(R)/\langle y^* \rangle) = 1$, but $x^{n-1} \in (\mmm^{n+1}: y) \setminus \mmm^n$ for each $n \geq 2$, i.e. $y$ is
*not* supercial.
**Edit: this example is incorrect (see the comments).** Let $R$ be the localization at the origin of $\kk[x,y,z,w]/\langle x^2 - z^3 , xy - w^3 \rangle$. Then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*,w^*]/\langle (x^*)^2, x^*y^* \rangle$. Consequently, $\dim(\gr_\mmm(R)/\langle y^* \rangle) = 1$, but $xz^{n-2} \in (\mmm^{n+1}: y) \setminus \mmm^n$ for each $n \geq 2$, so that $y$ is not superficial.
- An example where Condition 1 is satisfied but Condition 2 fails: Let $R$ be the localization at the origin of $\kk[x,y,z,w]/\langle xz-y^3,yz-x^4,z^2-x^3y^2 \rangle$. Then $\gr_\mmm(R) \cong \kk[x^*,y^*,z^*,w^*]/\langle (z^*)^2, y^*z^*, x^*z^*, (y^*)^4 \rangle$. Consequently, $\dim(\gr_\mmm(R)/\langle x^* \rangle) = 1$, but $zw^{n-2} \in (\mmm^{n+1}: x) \setminus \mmm^n$ for each $n \geq 2$, so that $x$ is not superficial.

Now we prove the theorem following arguments from the proof of the existence of superficial elements in Volume 2 of Zariski-Samuel (Chapter VIII, Section 8, Lemma 5).

Let $\ppp_1, \ldots, \ppp_k$ be the prime ideals of $G := \gr_\mmm(R)$ associated to the zero ideal. We may assume that $\ppp_j$ contains the maximal ideal $G_+ := \oplus_{i \geq 1}\mmm^i/\mmm^{i+1}$ if and only if $j > k'$, where $k' \leq k$. Since $G$ satisfies the two conditions mentioned in the beginning, each of $\ppp_1, \ldots, \ppp_{k'}$ is a *minimal* prime ideal of dimension $d$ associated to zero. In particular, $f^* \not\in \ppp_j$ for $j = 1, \ldots, k'$.

Let $\qqq_j$ be a primary component of the zero ideal of $G$ corresponding to $\ppp_j$. For each $j > k'$, $\qqq_j$ contains a power of $G_+$. Pick $c$ such that
$$G_+^c \subseteq \bigcap_{j > k'} \qqq_j$$
Pick $n \geq c$ and $g \in (\mmm^{n+\alpha}: \langle f \rangle) \cap \mmm^c$. It suffices to show that $g \in \mmm^n$. Indeed, if $g \not\in \mmm^n$, then $g^*f^* = 0 \in G$, where $g^*$ is the initial form of $g$. Since $f^* \not\in \ppp_j$, $j = 1, \ldots, k'$, it follows that $g^* \in \qqq_j$, $j = 1, \ldots, k'$. On the other hand, $g^* \in G_+^c$ (since $g \in \mmm^c$). Consequently,
$$ g^* \in \bigcap_{j=1}^k \qqq_j = 0$$
But then $g = 0 \in \mmm^n$. This contradiction proves the theorem.