Let $\mathbb{P}$ be any probability measure concentrated on an interval $[-1, 1]$. Why is it that for $\varphi$, the standard normal density, the following holds?
$$ (\varphi * \mathbb{P}) (x) = \int_{-1}^{1}\phi(x-y) \, {\rm d} \mathbb{P}(y)\geq \frac{\exp(-x^2)}{e\sqrt{2\pi}} $$
I tried to show this for the Dirac measure concentrated at $-1, 1$ depending on the sign of $x$ but I don't see how the resulting $\varphi(x-1), \varphi(x+1)$ must satisfy the inequality. Furthermore, I tried to set up an argument along the lines of $\varphi$ is decreasing/increasing , hence the following inequalities hold... using the edge cases above. How do I make such an argument make sense?
We have the following relation: \begin{align*} (\varphi * \mathbb{P})(x) & = \int_{-1}^{1} \varphi(x - y) \ \mathrm{d} \mathbb{P}(y) = \frac{1}{\sqrt{2\pi}}\int_{-1}^{1} \exp\left( -\frac{(x-y)^2}{2} \right) \ \mathrm{d} \mathbb{P}(y) \\ &= \frac{\exp( -{x^2})}{\sqrt{2\pi}} \int_{-1}^{1} \exp\left( \frac{-y^2 + 2xy + x^2}{2} \right) \ \mathrm{d} \mathbb{P}(y). \end{align*}
Let us fix a value of $x$. We define $g(y) = 0.5(-y^2 + 2xy + x^2)$. It is straightforward to note that $g(y)$ is increasing for $y < x$ and decreasing for $y \geq x$. Thus, if $x \geq 1$, $g(y)$ is increasing for $[-1,1]$ and hence $g(y) \geq g(-1)$. Similarly, if $x \leq -1$, we have, $g(y) \geq g(1)$. Lastly, if $|x| \leq 1$, then $g(y)$ is increasing for $[-1, x)$ and decreasing for $(x,1]$. Consequently, $g(y) \geq \min\{g(-1), g(1)\}$. Note that we can consolidate all the cases as follows: \begin{align*} g(y) \geq \min\{g(-1), g(1)\} \geq \inf_{x \in \mathbb{R}} 0.5 \min\{ x^2 - 2x -1, x^2 + 2x -1\} \geq \inf_{x \in \mathbb{R}} 0.5 (x^2 - 2|x| -1) \geq -1. \end{align*} The last inequality is obtained using some elementary calculus. Thus, we have, \begin{align*} (\varphi * \mathbb{P})(x) &= \frac{\exp( -{x^2})}{\sqrt{2\pi}} \int_{-1}^{1} \exp\left( \frac{-y^2 + 2xy + x^2}{2} \right) \ \mathrm{d} \mathbb{P}(y) \\ & \geq \frac{\exp( -{x^2})}{\sqrt{2\pi}} \int_{-1}^{1} \exp\left( -1 \right) \ \mathrm{d} \mathbb{P}(y) \\ & \geq \frac{\exp( -{x^2})}{e\sqrt{2\pi}}. \end{align*}