Suppose that $\mu(dx) = \exp(-\psi(x)) \, \mathrm{dx}$ is a probability measure on $\mathbf{R}^d$.
For suitable functions $g \geqslant 0$, define
$$\text{Ent}(g) = \int \mu(dx) g(x) \log \left( \frac{g(x)}{\int \mu(d\bar{x}) g(\bar{x})} \right).$$
We say that $\mu$ satisfies $\text{LSI}(c)$ (logarithmic Sobolev inequality with constant $c$) if for all suitable functions $g$,
$$\text{Ent}(g^2) \leqslant 2c \int \mu(dx) |\nabla g (x) |^2 .$$
Roughly speaking, the more well-concentrated the measure $\mu$ is, the smaller $c$ can be taken. For example, if $\mu = \mathcal{N}(0, \sigma^2)$, then one can take $c = \sigma^2$ (I think).
Now, suppose that $\mu$ satisfies $\text{LSI}(c)$, and for $\lambda \geqslant 0, x_0 \in \mathbf{R}^d$, we define
$$\nu (dx) = \frac{1}{Z} \mu (dx) \cdot \exp \left( - \frac{\lambda}{2} |x - x_0|_2^2 \right),$$
where $Z$ is a normalisation constant.
For $\lambda > 0$, one might intuitively expect that $\nu$ will become more tightly-concentrated than $\mu$, and that this would be reflected in the log-Sobolev constant of $\nu$.
My questions are: i) is this intuition correct?, ii) if so, can it be proven?, and iii) if so, can the estimate be made quantitative? e.g. does it holds that $\nu$ satisfies $\text{LSI}(\tilde{c})$, where $\tilde{c} = \frac{c}{1 + \lambda c} \leqslant c$ (at a guess), or for some other explicit $\tilde{c} \leqslant c$?
