I want to prove that if X is such that $$P[|X-m_X|\geq t] \leq c_1 e^{-c_2t^2},$$ for $c_1, c_2$ positive constants, $t\geq 0$, then it holds that $$P[|X-E[X]|\geq t] \leq c_3 e^{-c_4t^2},$$ with $c_3=1+2c_1$ and $c_4=c_2/4$.
There is a proof of reverse direction here.
First, note that we may re-write the problem using auxiliary variable $Y=X-m_X$. Hence, we need to show,
$$ P[|Y|\geq t] \leq c_1 e^{-c_2t^2} \Rightarrow P[|Y-E[Y]|\geq t] \leq c_3 e^{-c_4t^2}.$$
A proof may be obtained using the properties of sub-Gaussian random variables. It can be shown that if $P[|Y|\geq t] \leq c_1 e^{-c_2t^2} $ then there exist finite $c_3,c_4$ such that $E[\exp(c_4Y^2)]\leq c_3$ (You may find the proof in Proposition 2.5.2 of R.Vershynin (2018) book). Hence, $$P[|Y|\geq t] \leq c_1 e^{-c_2t^2} \Rightarrow \exists c_3,c_4>0,\quad E[\exp(c_4Y^2)]\leq c_3.$$ By Markov inequality we may write $$P[|Y-E[Y]|\geq t] = P\left[\exp(c_4|Y-E[Y]|^2)\geq \exp(c_4t^2)\right] \leq e^{-c_4t^2}E\left[\exp(c_4|Y-E[Y]|^2)\right].$$ Then it can be easily seen that: $$e^{-c_4t^2}E\left[\exp(c_4|Y-E[Y]|^2)\right] \leq e^{-c_4t^2}E\left[\exp(c_4|Y|^2)\right],$$ which gives $$P[|Y-E[Y]|\geq t]\leq e^{-c_4t^2}E\left[\exp(c_4|Y|^2)\right]\leq c_3 e^{-c_4t^2}.$$
Vershynin, Roman, High-dimensional probability. An introduction with applications in data science, Cambridge Series in Statistical and Probabilistic Mathematics 47. Cambridge: Cambridge University Press (ISBN 978-1-108-41519-4/hbk; 978-1-108-23159-6/ebook). xiv, 284 p. (2018). ZBL1430.60005.
