Are Lp norms equivalent on certain space of functions? Here, functions are of course infinite dimensional. I know that if functions are allowed to have singularity and be non-zero everywhere, then Lp norms are certainly not equivlent.
If functions have support and no singularity, will things change? If I have to go further, giving a strick upperbound to the functions would be enough right?
If by equivalent you mean topologically equivalent, no. Consider $f(x)=\begin{cases} 1 & x \in [0,\epsilon] \\ 0 & \text{otherwise} \end{cases}$ then $\| f \|_{L^\infty}=1$ but $\| f \|_{L^1}=\epsilon$ which can be arbitrarily small. The non-smoothness of this $f$ is not essential, either.