Show that $$\int_Re^{|f(x)|}dx=1\implies \sup_{1\leq p<\infty}p^{-1}\|f\|_p<\infty$$
To be honest, I could not even prove that $f \in L^1$. I feel like it can be done by contradiction, but i am not sure what not being integrable really implies about $|f|$. I can intuitivelly tell $f$ must decrease reallly fast for $e^{|f(x)|} \in L^1$ but I dont know how to formalize it, or how to use it here.
Thank you
Something's up with that question. $|f(x)| \geq 0$, so $e^{|f(x)|} \ \geq e^0 = 1$. Thus $\int e^{|f(x)|} \geq \int 1 = \infty$.
As a nice result, the question as stated is technically true, since a contradiction implies anything.
Edit: OP proposes maybe we want to consider $-|f(x)|$. But then we need $|f(x)|$ large except on a set of finite measure, e.g. $|f(x)| > 1$ except on a set of finite measure, otherwise $e^{-|f(x)|} > e^{-1}$ on a set of infinite measure, which would give $\int e^{-|f(x)|} = \infty$. So $f(x)$ would not be in $L^p$ for any $p$.