Why does this inequality hold? [Sobolev Embedding]

Question

Assume that $2<p<\frac{N+2}{N-2}$ where we work in $\mathbb{R}^N.$ Then $p+1< q^{*} = \frac{2N}{N-2}.$ Thus by sobolev embedding we have that $H^1 \hookrightarrow L^{q}$ for all $q\leq q^{*}$ or in other words we have the estimate, $$\|f\|_{L^{q}} \leq C \,\|f\|_{H^1}.$$ I am trying to show that $\|f^p\|_{L^2}\leq C \,\|f\|^2_{H^1}$ for some constant $C>0.$ I wrote $\|f^p\|_{L^2} = \|f\|^2_{L^{2p}}$ but then this would require that $H^{1} \hookrightarrow L^{2p}$ which is not necessarily true. Any hints on how to prove this will be much appreciated.

Answer 1

So first, in $\mathbb{R}^N$, the Sobolev embedding works only for $q\in[2,q^*)$.

Then, as you say, $H^1 ⊂ L^{2p}$ will only work when $2p ∈ [2,q^*]$, or equivalently $$ p∈[1,\tfrac{N}{N-2}] $$ Sobolev embeddings are sharp, so when $p ∈ (\tfrac{N}{N-2},\tfrac{N+2}{N-2})$, there are counterexamples (take for example $\varphi$ bounded and compactly supported and $f(x) = \varphi(x)\,|x|^{\varepsilon-N/p}$), and so in these cases $f^p \notin L^2$.