Let $p \in [1,\infty)$ and $f \in L^p((0,1),\mathbb{R})$. I'm interested in non-trivial sufficient conditions which ensure $f \in L^{\infty}((0,1),\mathbb{R}).$
After thinking a bit about this, I believe that $f$ having compact support could be one such condition. However, I can't think of a reason why!
I would be happy if anyone could provide any references for the same.
No, no conditions on the support will give you any $L^\infty$ bound on your function. It i just saying that the function is zero for some regions, but it can still be as you want in other regions ... Example: $$ f(x) = \frac{1}{|x-1/2|^{1/{2p}}}\ \mathbf{1}_{[-1/4,3/4]}(x) $$ is in $L^p$ and compactly supported. You even have functions that are in $L^p$ for every $p<\infty$ (and that are compactly supported) but are not bounded, like $$ f(x) = \ln(|x-1/2|)\ \mathbf{1}_{[-1/4,3/4]}(x). $$
Usual non-trivial criteria to get boundedness of functions is using conditions on the regularity of the function (i.e. $L^p$ bounds for gradients). You should then look about "Sobolev embeddings" and "Gagliardo-Nirenberg inequalities"