We are in $(L^2[0,1],\|.\|_{L^2})$ with some basis $\{\phi_k\}$. For $f\in L^2$ define $f_k:=\langle f,\phi_k\rangle$. Furthermore assume some smoothness condition on $f$: $$ \sum_{k=1}^\infty k^{2s}f_k^2<1\,\,\,\text{for some s > 0 }. $$ From this it follows $$\sum_{M=k}^\infty f_k^2\leq M^{-2s}$$ for any $M\in\mathbf{N}$.
Now define M such that $$ \|f\|^2_{L^2}=\sum_{k=1}^\infty f_k^2= 2 M^{-2s} $$
I need to proof for some $\gamma>0$ that $$ \sum_{k=1}^\infty k^{-\gamma}f_k^2 \lesssim M^{-2s-\gamma} $$ Thank you for any suggestions.
What i tried is to define N as $$ \sum_{k=1}^Nk^{-\gamma} f_k^2=\sum_{k=N+1}^\infty k^{-\gamma}f_k^2 $$ Then the left hand side $\sum_{k=1}^\infty k^{-\gamma}f_k^2$ is smaller than $2N^{^-2s-\gamma}$ so it remains to show that N<M but i have no clue how to proof this.