Let $\mu , \nu$ be probability measures on $\Bbb R$ with symmetric densities with respect to the lebesgue measure.
Let us assume that $\mu ([-c, c]) \geq \nu ([-c,c])\ \forall c \geq 0$ (So $\mu$ puts more mass towards zero).
What I want to show (I don't even know whether it's true or not, but I think so) is:
Then we have $(\mu * \mathcal N_{0,t}) ([-c,c]) \geq (\nu * \mathcal N_{0,t}) ([-c,c])$, where $\mathcal N _{0,t}$ is the normal distribution with mean $0$ and variance $t>0$.
Let $f_\mu , f_\nu$ be the densities of $\mu$ and $ \nu$, respectivly. Let $\varphi_t (x) := \frac 1 {\sqrt{2\pi t}} e^{-\frac{x^2} {2t}}$. Because of symmetry it suffices to show that $$\int_0^c \int_{\Bbb R} (f_\mu (y) - f_\nu (y))\varphi_t (x-y) \text d y \text d x \geq 0$$ Let $h := f_\mu - f_\nu$. Then $$\int_0^c \int_{\Bbb R} h(y) \varphi_t (x-y)\text d y \text d x = \int_{\Bbb R} h(y)\int_0^c \varphi_t (x-y)\text d x \text d y\\ = \int_{\Bbb R} h(y) \int_{-y}^{c-y} \varphi_t (x)\text d x \text d y\\ =\int_0^\infty h(y) \int_{-y}^{c-y} \varphi_t (x)\text d x \text d y + \int_0^\infty h(y) \int_{y}^{c+y} \varphi_t (x)\text d x \text d y\\ =\int_0^\infty h(y) \int_{y-c}^{y} \varphi_t (x)\text d x \text d y + \int_0^\infty h(y) \int_{y}^{c+y} \varphi_t (x)\text d x \text d y\\ =\int_0^\infty h(y) \int_{y-c}^{y+c} \varphi_t (x)\text d x \text d y$$ Define $\omega (y) := \int_{y-c}^{y+c} \varphi_t (x)\text d x$ for $y\geq 0$. The function $\omega$ is continuously differentiable and non-increasing (i.e. $\omega ' \leq 0$). Furthermore define $H(z) := \int_0^z h(y) \text d y$ for $z\geq 0$.
Thus above equals by integration by parts $$\int_0^\infty h(y) \omega (y) \text d y = \lim_{z\to\infty} \int_0^z \frac{\text d}{\text d y} (H(y)\omega(y)) - \int_0^z \underbrace{H(y)}_{\geq 0}\underbrace{\omega ' (y) }_{\leq 0}\text d y\\ \geq \lim_{z\to\infty} H(z) \omega (z) \geq 0$$ since $H(z) \geq 0 $ for all $z\geq0$.