I am a book on variational methods in a chapter on the Thomas-Fermi theory has this Lemma without a full proof.
a) The function $W: \mathbb{R}^3 \rightarrow \mathbb{R}, W(x)=\frac{1}{|x|}, x\neq0$ can be written as \begin{equation} W=W_1 +W_2 \text{ where } W_j \in L^{p_j}(\mathbb{R}^3) \text{ s.t. } 1 \leq p_1 <3 < p_2 \leq \infty \end{equation} therefore \begin{equation} W_1(x)=\begin{cases} \frac{1}{|x|} & |x| \leq R, x \neq 0 \\ 0 & |x|>R>0 \end{cases} \end{equation}
b) \begin{equation} (f,g) \rightarrow \langle f,g\rangle \equiv \langle f, W*g \rangle_2 = \int_{\mathbb{R}^3} \frac{f(x)g(y)}{|x-y|} {\mathrm {d}}x dy \end{equation} gives the inner (scalar) product on $L^1 \cap L^{\frac{5}{3}}$. Therefore we can use Cauchy-Schwarz \begin{equation} |\langle f,g\rangle| \leq \|f\|_{q^1_2} \|W_1\|_{q^1_2}\|g\|_{q^1_3}+\|f\|_{q^2_1} \|W_2\|_{q^2_2} \|g\|_{q^2_3} \end{equation} holds for all $\{q^i_j : i=1,2, j=1,2,3\}$ with \begin{equation} 2= \sum^3_{j=1} \frac{1}{q^i_j}, i=1,2 \text{ and } q^i_1 , q^i_3 \in [1, \frac{5}{3}], q^1_2 <3 <q^2_2 \end{equation} Thus the inner product is continuous with respect to the norm $\| \cdot \|_{\frac{5}{3}} + \| \cdot \|_r , 1 \leq r <\frac{5}{3}$
From earlier in the book:
From $f_j \in L^{q_j}, j=1,2$ if follows that $f_1 *f_2 \in L^{q_3}$ where \begin{equation} 1+\frac{1}{q_3} = \frac{1}{q_1} + \frac{1}{q_2} \end{equation} and \begin{equation} \|f_1 * f_2\|_{q_3} \leq \|f_1\|_{q_1}\|f_2\|_{q_2} \text{ (Young's inequality)} \end{equation}
My attempt at a proof: a) I have no idea b) \begin{equation} \begin{aligned} |\langle f, W*g \rangle_2| &\le \|f\|_2 \|W*g\|_2 \\ &= \|f\|_2 \|(W_1+W_2)*g\|_2 \\ &\le \|f\|_2 (\|W_1*g\|_{p_1} +\|W_2*g\|_{p_2}) \text{ p as defined above} \\ &= \|f\|_2 \|W_1*g\|_{p_1} + \|f\|_2 \|W_2*g\|_{p_2} \\ &\le \|f\|_2 \|W_1\|_{q_2^1} \|g\|_{q_3^1} + \|f\|_2 \|W_2\|_{q_2^2}\|g\|_{q_3^2} \end{aligned} \end{equation} Where: \begin{equation} 1+ \frac{1}{p_1}=\frac{1}{q_2^1}+\frac{1}{q_3^1} \qquad 1+ \frac{1}{p_2}=\frac{1}{q_2^2}+\frac{1}{q_3^2} \end{equation} This leads to \begin{equation} \frac{4}{3}<\frac{1}{q_2^1}+\frac{1}{q_3^1} \le 2 \qquad 1 \le \frac{1}{q_2^2}+\frac{1}{q_3^2} < \frac{4}{3} \end{equation} Now this is obviously wrong but I am not sure how to proceed, so any help would be appreciated.