I would like to know whether $\|f\|_{L^2}\leq\mu(X)^{1/2}\|f\|_{L^1}$ for bounded functions. I can show an opposite relation given a finite measure space using Holder's Inequality, but I am interested in an inverse relation, even if there are constants that need to be added/multiplied.
I do not think that this question is a duplicate. I am aware that there are counterexamples for the general case where the function is not bounded, such as: $$f(x)=\frac{1}{\sqrt{x}}$$ but I have seen people mention that the inequality holds true for functions with no singularities.
Intuitively it does make sense to me but I am having trouble trying to prove it, and I could not find any reference to that in books.
Since $|f(x)|^2 \leq \|f\|_{L^\infty}\,|f(x)|$, $$ \|f\|_{L^2} = \left(\int |f(x)|^2\,\mathrm d x\right)^{1/2} \leq \left(\int \|f\|_{L^\infty}|f(x)|\,\mathrm d x\right)^{1/2} = \|f\|_{L^\infty}^{1/2}\, \|f\|_{L^1}^{1/2}. $$ In particular, $$ \|f\|_{L^2} \leq \|f\|_{L^\infty} + \|f\|_{L^1}. $$