Let $M$ denote the median of a function $f(X)$ that is Lipschitz continuous with $\left \| f \right \|_{Lip}=1$. I am trying to show that if $\left \| f(X)-M \right \|_{\psi_{2}}\leq C$, then $\left \| f(X)-\mathbb{E}f(X) \right \|_{\psi_{2}}\leq C$, where $\psi_{2}$ is the sub-gaussian norm. I understand that the inverse is true due to the fact that the median minimizes the $L_{1}$ norm which is bounded by the sub-gaussian norm.
For this direction I thought about using
\begin{align} \left \| f(X)-\mathbb{E}f(X) \right \|_{\psi_{2}}=\left \| f(X)-M+M-\mathbb{E}f(X) \right \|_{\psi_{2}}\\ \leq\left \| f(X)-M \right \|_{\psi_{2}}+\left \| M-\mathbb{E}f(X) \right \|_{\psi_{2}}\\ \leq C+\left \| M-\mathbb{E}f(X) \right \|_{\psi_{2}} \end{align}
But I don't know how to proceed.