# Show that elements of $H^{\frac{n}{2} + 1}(\mathbb{R}^n)$ need not be Lipschitz

Say a function $u : \mathbb{R}^n \to \mathbb{C}$ is $\alpha$-Hölder continuous, $0 < \alpha <1$, if $u$ is bounded and for all $x, y \in \mathbb{R}^n$ with $|y| \le 1$ $$|u(x +y) - u(x)| \le M |y|^\alpha, \qquad \text{some } M > 0.$$ If the same estimate holds instead for $\alpha = 1$, then we say $u$ is Lipschitz.

Denote by $H^s(\mathbb{R}^n)$, $s \in \mathbb{R}$, the standard Sobolev subspace of tempered distributions on $\mathbb{R}^n$,

$$H^s := \{u \in \mathcal{S}'(\mathbb{R}^n): \langle \xi \rangle^{s}\widehat{u} (\xi) \in L^2(\mathbb{R}^n) \},$$

where $\langle \xi \rangle := (1 + |\xi|^2)^{\frac{1}{2}}$ and "hat" denotes Fourier transform.

A standard Sobolev embedding result is that if $u \in H^{\frac{n}{2} + \alpha }(\mathbb{R}^n)$, $0 < \alpha <1$, then $u$ is $\alpha$-Hölder continuous. However, it is also asserted, but not proved, in chapter 4 of Taylor Vol. I that functions in $H^{\frac{n}{2} + 1}(\mathbb{R}^n)$ need not be Lipschitz.

I would like to show that functions in $H^{\frac{n}{2} + 1}(\mathbb{R}^n)$ need not be Lipschitz.

One strategy is to use fact that $u$ is Lipschitz if and only if $u$ has bounded partial derivatives. So we seek a function $u$ such that $\langle \xi \rangle^{\frac{n}{2} + 1} \widehat{u}(\xi) \in L^2$ but $|\nabla u|$ is unbounded.

Another strategy is to use Fourier inversion, which can be shown to hold for elements of $H^{\frac{n}{2} + 1}(\mathbb{R}^n).$ That is we seek $u \in H^{\frac{n}{2} + 1}(\mathbb{R}^n)$ for which

$$\sup_{0<|y| < 1}\frac{|u(x + y) - u(x)|}{|y|} =\sup_{0<|y| < 1} (2\pi)^{-\frac{n}{2}} |y|^{-1} \left| \int \widehat{u}(\xi) e^{i x \cdot \xi }(e^{i y \cdot \xi} - 1)d\xi\right| = \infty$$

However, I have yet to use use either of these strategies successfully. Hints or solutions are greatly appreciated!

Begin with an unbounded function in $H^{n/2}$, the critical case of Sobolev embedding theorem. The basic idea is that $\log |x|$ is very close to being in $H^{n/2}$, as its $(n/2)$ derivative is homogeneous of degree $-n/2$ which makes it very close to being square integrable. So, anything that has milder singularity than the logarithm will do, for example $(-\log|x|)^p$ with $0<p<1$. Such an example is considered here among other places. Here and below I'm focusing on the singularity at $0$, ignoring the behavior at infinity (which can be removed by multiplying with a smooth cutoff function).
To move up one order of smoothness, one could try to integrate but it's easier to multiply by a linear function; this has a similar effect on tempering the singularity at $0$. So, let $$u(x) = x_1(-\log|x|)^p,\quad 0<p<1$$ This is evidently not Lipschitz, as $u(te_1)/t \to \infty$ as $t\to 0$ (where $e_1$ is the first standard basis vector). On the other hand, $\nabla u$ is in $H^{n/2}$ and so $u\in H^{n/2+1}$. Indeed, $$\nabla u(x) = (-\log|x|)^p e_1 + x_1\nabla ((-\log|x|)^p) = (-\log|x|)^p e_1 -p \frac{x_1x}{|x|^2} (-\log|x|)^{p-1}$$ As previously noted, the first term $(-\log|x|)^p e_1$ is in $H^{n/2}$. The singularity of the second one is even milder due to smaller exponent of the logarithm.