Let $f$ be a "sufficiently smooth" probability density on $\mathbb R^n$. For $\theta=(w,c) \in \Theta := \{(w,c) \in \mathbb R^{n+1} \mid w \ne 0\}$ and $t \ge 0$, define
$$ F_t(\theta) := \int_{\mathbb R^n} H(t\|w\| - (x^\top w-c))f(x)\,dx, $$
where $H:\mathbb R \to \mathbb R$ is the Heaviside / unit-step function.
Question. Is it true that for every $\theta \in \Theta$, $F_t$ is locally convex at $\theta$, i.e., there exists an open neigborhood $U$ of $\theta$ such that $Hess(F_t) \succeq 0$ on $U \cap \Theta$ ?
N.B.: Global convexity of $F_t$ is probably hopeless because $H$ is not convex.
Observations
Partial derivative w.r.t $c$: $$ \dfrac{\partial F_t(\theta)}{\partial c} = \int_{\mathbb R^n}\delta(t\|w\|-(x^\top w-c))f(x) = \int_{x^\top w = c+t\|w\|}f(x)\,d\sigma_{n-1}(x) =: R[f](w,c+t\|w\|), $$ where $\delta$ is the Dirac delta distribution, and $R[f](w,b)$ is the Radon transform of $f$, evaluated at $(w,b)$.
Partial derivative w.r.t $j$th coordinate of $w$: $$ \begin{split} \dfrac{\partial F(\theta)}{\partial w_j} &= \frac{tw_j}{\|w\|}\int_{\mathbb R^n}\delta(t\|w\|-(x^\top w - c))f(x)\,dx - \int_{\mathbb R^n}\delta(t\|w\|-(x^\top w-c))x_jf(x)\,dx\\ & = \frac{tw_j}{\|w\|}R[f](w,c+t\|w\|) - R_j[f](w,c+t\|w\|) = \left(\frac{tw_j}{\|w\|}R-R_j\right)[f](w,c+t\|w\|), \end{split} $$ where $R_j[f]:(w,b) \mapsto \int_{\mathbb R^n}\delta(b-x^\top w)x_jf(x)\,dx$ is the Radon transform of the function $x=(x_1,\ldots,x_n) \mapsto x_j f(x)$