The Hardy-Littlewood-Sobolev inequality is the statement that there is a $C>0$ such that
$$
\tag{HLS}
\lVert f\ast \lvert\cdot\rvert^{-\alpha}\rVert_p\le C\lVert f\rVert_q, $$
for all $f\in L^q(\mathbb R^d)$, where the convolution is defined as
$$(f\ast \lvert\cdot\rvert^{-\alpha} )(x)= \int_{\mathbb R^d} \frac{f(y)}{\lvert x-y\rvert^{\alpha}}\, dy, $$
and the parameters satisfy the conditions
$$\tag{1} 0<\alpha<d, \quad \frac1p-\frac1 q+1=\frac\alpha d.$$
Now let us define the Japanese bracket $$\langle x \rangle:= \sqrt{1+ \lvert x\rvert^2},\qquad x\in\mathbb R^d,$$
which is a non-homogeneous version of $\lvert\cdot\rvert$. The function $\langle x\rangle^{-\alpha}$ has the exact same decay at $\lvert x \rvert \to \infty$ of its homogeneous counterpart $\lvert x \rvert^{-\alpha}$, but $\langle x \rangle^{-\alpha}$ is not singular at $x=0$. By the obvious estimate $\langle x\rangle^{-\alpha}\le \lvert x\rvert^{-\alpha}$, (HLS) immediately implies its non-homogeneous version
$$\tag{n-HLS}
\lVert f\ast \langle\cdot\rangle^{-\alpha}\rVert_p\le C\lVert f\rVert_q, $$
under the assumptions (1).
Question. Is there a direct proof of (n-HLS) that is simpler than a proof of (HLS)?
Remark 1. The inequality (n-HLS) actually holds for $\frac1 p – \frac1q +1 \le \frac{\alpha}{d}$. However, the non-endpoint case $\frac1 p – \frac1q +1 < \frac{\alpha}{d}$ can be immediately proved by an application of the Young inequality for convolutions. Thus, the present question is concerned solely with the endpoint case (1).
In the following, the letter $C$ will always denote an irrelevant positive constant, whose value may change from line to line.
Remark 2. There are many proofs of (HLS). One that I know uses the Hardy-Littlewood maximal function. By decomposing the ball $B(x, r)$ into dyadic annuli, we obtain the local estimate
$$\tag{2}\left\lvert \int_{B(x, r)}\frac{f(y)}{\lvert x-y \rvert^\alpha}dy\right\rvert \le r^{d-\alpha}\sum_{j=0}^\infty 2^{\alpha j – dj} Mf(x)= C_{d, \alpha}r^{d-\alpha}Mf(x).$$
This constant $C_{d,\alpha}$ equals $\sum 2^{(\alpha – d)j}$, which is finite because $\alpha<d$. Here $Mf$ denotes the Hardy-Littlewood maximal function
$$
Mf(x):=\sup \frac1{\lvert B\rvert} \int_B \lvert f(y)\rvert\, dy,$$ where the sup is taken over all balls $B$ containing $x$.
We then estimate the tail of the convolution via the Hölder inequality;
$$\tag{3}
\left\lvert \int_{\mathbb R^d\setminus B(x, r)}\frac{f(y)}{\lvert x-y \rvert^\alpha}dy\right\rvert\le C r^{\frac{d}{q'}-\alpha} \lVert f\rVert_q.$$
Combining (2) and (3) gives the pointwise bound
$$
\tag{4} \lvert f\ast \lvert\cdot\rvert^{-\alpha}(x)\rvert \le C\left( r^{d-\alpha} Mf(x) + r^{\frac{d}{q'}-\alpha} \lVert f\rVert_q\right).$$
Choosing the $r$ that minimises the right-hand side, then integrating and applying the Hardy-Littewood maximal estimate $\lVert Mf\rVert_q\le C\lVert f \rVert_q$, we obtain (HLS).
This procedure will of course work if $\lvert\cdot\rvert^{-\alpha}$ is replaced by $\langle\cdot\rangle^{-\alpha}$. However, I think that there shouldn't be the need for the careful dyadic analysis of equation (2). Indeed, $\langle x-y\rangle^{-\alpha}$ is not singular at $y=x$. This is why I am asking for a simpler proof.
Best Answer
In the given range $$\tag{1}\frac1p + 1 = \frac1q+\frac\alpha d, $$ there is no hope for a simpler proof in the non-homogeneous case.
To explain, let me consider the weak Young inequality $$\tag{2} \lVert f\ast g \rVert_p \le C\lVert f\rVert_{q_1} \lVert g\rVert_{q_2,\infty}, \quad \frac1p+1 = \frac1{q_1}+\frac1{q_2},$$ where $$\tag{3}\lVert g\rVert_{q_2,\infty} = \sup_{t>0} t \lvert \{ \lvert g\rvert >t\}\rvert^\frac1{q_2}.$$ We want to prove (n-HLS) under the assumption (1), which, according to (2) with $g=\langle\cdot\rangle^{-\alpha}$, follows from $$\tag{4} \lVert \langle \cdot \rangle^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty.$$ Now, it is clear that there is no simpler way to prove (4) than to estimate pointwise $\langle x \rangle^{-\alpha}\le \lvert x \rvert^{-\alpha}$, so that $$\lVert \langle\cdot \rangle^{-\alpha}\rVert_{\frac d \alpha, \infty} \le \ \lVert \lvert\cdot \rvert^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty,$$ which completes the proof. Thus the main step is the proof that $\lVert \lvert\cdot \rvert^{-\alpha}\rVert_{\frac d \alpha, \infty}<\infty$, which yields (HLS) and (n-HLS) essentially at the same time.
No proof can rely on anything simpler than this.
Let me remark that the mathematics behind all this is essentially the Marcinkiewicz interpolation theorem. I have consulted the blog post of Terry Tao https://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/; the weak Young inequality is Exercise 44.
In the proof that I gave in the main question above, the weak-Lp computations are there. They are just hidden behind the Hardy-Littlewood maximal estimate $\lVert Mf\rVert_p\le C \lVert f\rVert_p$, which indeed is proven via the interpolation theorem of Marcinkiewicz.