Could you please help me solve the following problem:
Is it true that for any $\varepsilon>0$ there exist a $\delta=\delta(\varepsilon)>0$ (depends on $\varepsilon$) such that $$ \int_{A}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx<\delta $$ implies $$ \int_{A}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-t)^2}{2t}}dx<\varepsilon $$ ?
Speaking in measure terms, lets denote $\mathbb{P}(A)=\int_{A}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx$. Is it true that when $\mathbb{P}(A)$ is small, then $\mathbb{P}(\tilde A)$ is small too? Here $\tilde A$ is a set $A$ shifted by one unit of variance.
This is a particular case of a larger problem that I can't deal with. Would be grateful for any tips!
I have understood how to solve it: let us denote $p(x)=\frac{1}{\sqrt{2\pi t}}\exp(-\frac{x^2}{2t})$ - the density of a centered normal variable with density $t$. If $$ \mathbb{P}(A)=\int_Ap(x)dx=\int_A \frac{1}{\sqrt{2\pi t}}\exp(-\frac{x^2}{2t})dx<\delta $$ then $$ \int_A \frac{1}{\sqrt{2\pi t}}\exp(-\frac{(x-t)^2}{2t})dx=\int_A \exp(-\frac{t}{2}+x)p(x)dx=\mathbb{E}(\exp(-\frac{t}{2}+\xi)\cdot I_A) $$ where $I_A$ - is an indicator of set $A$. Using Cauchy inequality $$ |\mathbb{E}(\xi\cdot \eta)|\le \sqrt{E\xi^2}\sqrt{E\eta^2} $$ we get $$ \mathbb{E}(\exp(-\frac{t}{2}+\xi)\cdot I_A)\le \sqrt{\mathbb{E}\exp(-t+2\xi)}\sqrt{\mathbb{E}(I_A)} $$
I guess that $\mathbb{E}\exp(-t+2\xi)$ should be bounded and $\sqrt{\mathbb{E}(I_A)}=\sqrt{\mathbb{P}(A)}<\delta^{\frac 12}$