Lieb, Seiringer: Stability of Matter in Quantum Mechanics equation 2.2.7 In page 27 of "Stability of Matter in Quantum Mechanics" by Lieb and Seiringer, it states:
An application of Hölder's inequality to (2.2.4) yields, for any potential $V\in L^{d/2}(\mathbb{R}^d),\ d\geq3$,
$$T_\psi\geq S_d\|\psi\|^2_{2d/(d-2)}\geq S_d(\psi,|V|\psi)\|V\|^{-1}_{d/2}.\qquad\qquad (2.2.7)$$
The inequality (2.2.4) is:
$$2T_\psi=\|\nabla\psi\|^2_2=\int_{\mathbb{R}^d}|\nabla\psi(x)|^2dx\geq S_d\|\psi\|^2_{2d/(d-2)}.$$
I don't see how Hölder's inequality leads to (2.2.7). The Hölder conjugate of $d/2$ is $d/(d-2)$ but inequality (2.2.4) deals with $2d/(d-2)$-norm. So, how do I get from (2.2.4) to (2.2.7)?
 A: The proof reduces to proving that $\langle\psi,V\psi\rangle_2\leq \lVert\psi\rVert_{\frac{2d}{d-2}}^2\lVert V\rVert_{\frac{d}{2}}$.
We can rewrite the scalar product as
$$\langle\psi,V\psi\rangle_2=\int_{\mathbb{R}^d}\bar{\psi}(x)V(x)\psi(x)\mathrm{d}x \leq \int_{\mathbb{R}^d} \lvert \bar{\psi}(x)V(x)\psi(x)\rvert \mathrm{d}x=\lVert \bar{\psi}V\psi\rVert_1\; .$$
Now, the Hölder inequality for three functions gives
$\lVert fgh\rVert_p\leq \lVert f\rVert_q \lVert g\rVert_r \lVert h\rVert_s$, with $p^{-1}=q^{-1}+r^{-1}+s^{-1}$. Since you want to bound both $\psi$ and $\bar{\psi}$ with the norm $\frac{2d}{d-2}$, you get
$$1=\frac{d-2}{2d}+\frac{d-2}{2d}+r^{-1}\; ,$$
where $r$ is the resulting index for the norm of $V$. However, $\frac{d-2}{d}=1-\frac{2}{d}$, hence it follows that $r^{-1}=\frac{2}{d}$, and thus $r=\frac{d}{2}$.
Therefore,
$$\langle\psi,V\psi\rangle_2\leq \lVert\psi\rVert_{\frac{2d}{d-2}}^2\lVert V\rVert_{\frac{d}{2}}$$
by Hölder (for three functions), as desired.
Let me also remark that for this application of Hölder's inequality, the restriction $d\geq 3$ is not necessary. What fails to be true if $d<3$, is the inequality $2T_{\psi}\geq S_d \lVert\psi\rVert_{\frac{2d}{d-2}}^2$ since it is a consequence of Sobolev's embedding theorem.
