This is a follow up to the problem posted here.
Show that for any $\epsilon>0$, there exists a constant $c_\epsilon$ such that $$|\psi(\textbf{x})|\le c_\epsilon\|\psi\|+\epsilon\|\Delta\psi\|$$ for all $\psi\in D(\Delta)$.
Here we define $\Delta\psi:=-\mathcal{F}^{-1}(|\textbf{k}|^2\hat\psi(\textbf{k}))$ where $\mathcal{F}^{-1}$ denotes the inverse fourier transform operator. One possibly useful observation is the unitarity of the inverse fourier transform. This gives $$\|\Delta\psi\|=\big\|-\mathcal{F}^{-1}(|\textbf{k}|^2\hat\psi(\textbf{k}))\big\|=\big\||\textbf{k}|^2\hat\psi(\textbf{k})\big\|.$$ Now, presumably we want to make use of the result in the previous post so we start from $$|\psi(\textbf{x})|\le c_1\|\psi\|+c_2\big\||\textbf{k}|^{9/5}|\hat\psi(\textbf{k})|\big\|$$ but where does this $\epsilon$ get introduced?
As you probably noticed already, $\lvert k\rvert^{9/5}\leq 1+\lvert k \rvert^2$, so that the inequality from the previous post implies $$ \lVert \psi\rVert_\infty\leq (1+c_1)\lVert \psi\rVert_2+c_2\lVert \Delta \psi\rVert_2. $$ Now the question is, how can we make the constant $c_2$ arbitrarily small (at the expense of a large constant in front of $\lVert \psi\rVert_2$). The answer to that is to use the symmetries of the left side:
If you replace $\psi$ by $\psi_\lambda$ given by $\psi_\lambda(x)=\psi(\lambda x)$, then the left side stays unchanged. On the right side you get $\lVert \psi_\lambda\rVert_2=\lambda^{-n/2}\lVert \psi\rVert_2$ and $\lVert \Delta \psi_\lambda\rVert_2=\lambda^{2-n/2}\lVert \Delta \psi\rVert_2$. Now you can choose $\lambda$ sufficiently small to get $$ \lVert \psi\rVert_\infty\leq c_\epsilon\lVert \psi\rVert_2+\epsilon\lVert \Delta\psi\rVert_2. $$