Let $r>2$ and let $r*$ be its dual coefficient, i.e., \begin{equation} \frac{1}{r}+\frac{1}{r*}=1. \end{equation} Let $a>0$ and $T>0$.
Does there exists a continuous, $T$-periodic function $f:[0,T]\to\mathbb{R}$ such that \begin{equation} \|f\|_{L^2(0,T)}>\|f-af'\|_{L^{r*}(0,T)}\|f\|_{L^r(0,T)}? \end{equation} Here, prime denotes the weak derivative.
Clearly, there can be no such function for $r=2$ by the polarization identity and the orthogonality of $f$ and $f'$.
I would very much appreciate any ideas.