Consider the ODE $ f''=f^{-\alpha} $, where $ \alpha>1 $ and $ f>0 $ in $ \mathbb{R} $. Assume that for $ [f]_{\frac{2}{\alpha+1}}\leq A $, where $ A>0 $ is a constant and $$ [f]_{\frac{2}{\alpha+1}}=\sup_{x,y\in\mathbb{R}}\frac{|f(x)-f(y)|}{|x-y|^{\frac{2}{\alpha+1}}}.\quad(*) $$ I want to ask if for any $ \epsilon>0 $, then there is a solution $ f $ such that $ \inf_{\mathbb{R}}f<\epsilon $.

Since $ f''>0 $, by simple analysis, we see that $ f(t)\to+\infty $ as $ t\to\infty $. I guess that by the Hölde assumption (*) will ensure that $ f $ cannot be very small. However I do not know how to go on. Can you give me some hints or references?

Consider the ODE $ f''=f^{-\alpha} $, where $ \alpha>1 $ and $ f>0 $ in $ \mathbb{R} $. Assume that for $ [f]_{\frac{2}{\alpha+1}}\leq A $, where $ A>0 $ is a constant and $$ [f]_{\frac{2}{\alpha+1}}=\sup_{x,y\in\mathbb{R}}\frac{|f(x)-f(y)|}{|x-y|^{\frac{2}{\alpha+1}}}.\label{1}\tag{$*$} $$ I want to ask if for any $ \epsilon>0 $, then there is a solution $ f $ such that $ \inf_{\mathbb{R}}f<\epsilon $.

Since $ f''>0 $, by simple analysis, we see that $ f(t)\to+\infty $ as $ t\to\infty $. I guess that by the Hölder assumption \eqref{1} will ensure that $ f $ cannot be very small. However I do not know how to go on. Can you give me some hints or references?