I encounter the following claim in one paper:

If $(-\Delta)^{\frac14}u\in L^{2,\infty}(\mathbb{R})$, then $u\in BMO(\mathbb{R})$. Equivalently, if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u\in L^{2,1}(\mathbb{R})$. Here $L^{2,\infty}$ and $L^{2,1}$ are Lorentz space and $\mathcal{H}$ is the Hardy space.

I do not know how to show this fact. My knowledge of Riesz potential tells me if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^2(\mathbb{R})$, but why does it lie in the smaller space $L^{2,1}$? On the other hand, if $(-\Delta)^{\frac14}u\in L^2(\mathbb{R})$, then $u\in BMO$. However, this claim says we actually just need $(-\Delta)^{\frac14} u\in L^{2,\infty}$.

The paper says the first half of the claim is contained in the paper: Adams, D. R. (1975). A note on Riesz potentials. Duke Mathematical Journal. I read Adams' paper and could not figure out why.

I encounter the following claim in one paper:

If $(-\Delta)^{\frac14}u\in L^{2,\infty}(\mathbb{R})$, then $u\in BMO(\mathbb{R})$. Equivalently in its dual version, if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^{2,1}(\mathbb{R})$. Here $L^{2,\infty}$ and $L^{2,1}$ are Lorentz space and $\mathcal{H}$ is the Hardy space.

I do not know how to show this fact. My knowledge of Riesz potential tells me if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^2(\mathbb{R})$, but why does it lie in the smaller space $L^{2,1}$? On the other hand, if $(-\Delta)^{\frac14}u\in L^2(\mathbb{R})$, then $u\in BMO$. However, this claim says we actually just need $(-\Delta)^{\frac14} u\in L^{2,\infty}$.

The paper says the first half of the claim is contained in the paper: Adams, D. R. (1975). A note on Riesz potentials. Duke Mathematical Journal. I read Adams' paper and could not figure out why.

I encounter the following claim in one paper:

If $(-\Delta)^{\frac14}u\in L^{2,\infty}(\mathbb{R})$, then $u\in BMO(\mathbb{R})$. Equivalently, if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u\in L^{2,1}(\mathbb{R})$. Here $L^{2,\infty}$ and $L^{2,1}$ are Lorentz space and $\mathcal{H}$ is the hardy space.

I do not know how to show this fact. My knowledge of Riesz potential tells me if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^2(\mathbb{R})$, but why does it lie in the smaller space $L^{2,1}$?

The paper says the first half of the claim is contained in the paper: Adams, D. R. (1975). A note on riesz potentials. Duke Mathematical Journal. I read Adams' paper and could not figure out why.

I encounter the following claim in one paper:

If $(-\Delta)^{\frac14}u\in L^{2,\infty}(\mathbb{R})$, then $u\in BMO(\mathbb{R})$. Equivalently, if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u\in L^{2,1}(\mathbb{R})$. Here $L^{2,\infty}$ and $L^{2,1}$ are Lorentz space and $\mathcal{H}$ is the Hardy space.

I do not know how to show this fact. My knowledge of Riesz potential tells me if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^2(\mathbb{R})$, but why does it lie in the smaller space $L^{2,1}$? On the other hand, if $(-\Delta)^{\frac14}u\in L^2(\mathbb{R})$, then $u\in BMO$. However, this claim says we actually just need $(-\Delta)^{\frac14} u\in L^{2,\infty}$.

The paper says the first half of the claim is contained in the paper: Adams, D. R. (1975). A note on Riesz potentials. Duke Mathematical Journal. I read Adams' paper and could not figure out why.