# Smooth cut-off in homogeneous Besov space

Given a Littlewood-Paley decomposition

$$1 = \chi(\xi) + \sum_{j \geq 0}\varphi(2^{-j} \xi), \quad \xi \in \mathbb R^n$$

where $$\chi$$ is smooth, supported on a ball, and $$\varphi$$ is smooth, supported on an annulus, let's ​consider the homogeneous Besov space

\begin{align*} \dot{B}^{s}_{p,r}(\mathbb R^n) = \{u \in S'(\mathbb R^n) : &\lim \limits_{n \to +\infty}||\chi(2^nD)u||_{L^{\infty}} = 0, \\ &||u||_{\dot{B}^{s}_{p,r}} = ||(2^{js}||\varphi(2^{-j}D)u||_{L^p})_{l \in \mathbb Z}||_{\ell^r} < +\infty\} \end{align*}

For $$s > 0$$ and $$p, r < \infty$$, it is shown in G. Leoni "A First Course in Sobolev Spaces" (Ch. 17) that convolution with an approximate identity converges in $$\dot{B}^{s}_{p,r}$$ to the original function.

Hence, $$C^{\infty} \cap \dot{B}^{s}_{p,r}$$ is a dense subset of $$\dot{B}^{s}_{p,r}$$.

Is it possible to obtain density of $$C_c^{\infty} \cap \dot{B}^{s}_{p,r}$$ by considering a smooth cut-off of an element in $$C^{\infty} \cap \dot{B}^{s}_{p,r}$$ ? In the inhomogeneous case, it is definitely possible.

I'm interested in approximating elements (by functions which are Schwartz or compactly supported) in intersections such as $$\dot{B}^{s}_{p,r} \cap L^q$$. If the method of convolution + cut-off works in $$\dot{B}^{s}_{p,r}$$, then it will work as well in $$\dot{B}^{s}_{p,r} \cap L^q$$.

I have many tools (Bernstein inequality, Embeddings, Bony Decomposition), but I cannot prove it. We may need to assume $$s < n/p$$.

## 1 Answer

I think I've come up with a proof.

Assume that $$u \in \dot{B}^{s}_{p,r}(\mathbb R^d)$$ with $$s < d/p$$ and $$p,r \in [1,+\infty]$$. Write $$\dot{\Delta}_j = \varphi(2^{-j}D)$$, $$j \in \mathbb Z$$.

Approximate identity convergence

Let $$(k_{\lambda})_{\lambda > 0} \subset S(\mathbb R^d)$$ be any approximate identity that lies in Schwartz space, i.e.

1. $$\int_{\mathbb R^d} k_{\lambda} = 1$$
2. $$\sup_{\lambda > 0} ||k_{\lambda}||_{L^1} < +\infty$$
3. $$\forall \delta > 0$$ : $$\lim \limits_{\lambda \to \infty} \int_{|x| \geq \delta}||k_{\lambda}||_{L^1} = 0$$

Trivially,

$$||\chi(2^nD)(u * k_{\lambda})||_{L^{\infty}} = ||(\chi(2^nD)u) * k_{\lambda}||_{L^{\infty}} \leq ||\chi(2^nD)u||_{L^{\infty}} \cdot ||k_{\lambda}||_{L^1}$$

Using Young-Convolution inequality, $$2^{js}||\dot{\Delta}_j(u * k_{\lambda})||_{L^p} = 2^{js}||(\dot{\Delta}_ju) * k_{\lambda}||_{L^p} \leq 2^{js}||\dot{\Delta}_ju||_{L^p} \cdot ||k_{\lambda}||_{L^1}$$

Hence, $$u * k_{\lambda} \in \dot{B}^{s}_{p,r}(\mathbb R^d)$$ and it is uniformly bounded in $$\lambda$$. By the Fatou property, the limit (along some subsequence) as $$\lambda \to +\infty$$ exists in $$\dot{B}^{s}_{p,r}(\mathbb R^d)$$.

Since $$\dot{B}^{s}_{p,r}(\mathbb R^d)$$ is embedded into $$S'(\mathbb R^d)$$ (which is Hausdorff) and $$u * k_{\lambda}$$ converges to $$u * \delta = u$$ in $$S'(\mathbb R^d)$$, the limit in $$\dot{B}^{s}_{p,r}(\mathbb R^d)$$ must be $$u$$.

Smooth cut-off convergence

If $$f$$ is Schwartz, then $$||u \cdot f||_{\dot{B}^{s}_{p,r}} \leq C [ ||f||_{L^{\infty}} + ||f||_{\dot{B}^{d/p}_{p,\infty}} + ||f||_{\dot{B}^{0}_{\infty,\infty}} ] \cdot ||u||_{\dot{B}^{s}_{p,r}}$$ thanks to paraproduct continuity and the embedding $$\dot{B}^{s}_{p,r} \subset \dot{B}^{s-d/p}_{\infty,r}$$.

Moreover, $$||f||_{\dot{B}^{0}_{\infty,\infty}} \leq C ||f||_{L^{\infty}}$$

Now, we need to study $$||f||_{\dot{B}^{d/p}_{p,\infty}} = \sup_{j \in \mathbb Z} (2^j)^{d/p} ||\dot{\Delta}_j f||_{L^p}$$ We must use something which looks like Bernstein inequality, but with a fractional derivative $$\partial^{d/p}$$ (defined by means of the Fourier transform). I think we can prove $$(2^j)^{d/p} ||\dot{\Delta}_j f||_{L^p} \leq C ||\partial^{d/p} \dot{\Delta}_j f||_{L^p} \leq C ||\partial^{d/p} f||_{L^p}$$

If $$f$$ is a smooth cut-off, $$f(x) = \chi(x/R)$$ with $$0 \leq \chi \leq 1$$, $$\chi \in C^{\infty}_c(\mathbb R^d)$$, then $$||u \cdot \chi(x/R)||_{\dot{B}^{s}_{p,r}} \leq C [ ||\chi||_{L^{\infty}} + ||\partial^{d/p} \chi||_{L^p} ] \cdot ||u||_{\dot{B}^{s}_{p,r}}$$ so it is uniformly bounded for $$R > 0$$. Thanks to Fatou property, the limit (along some subsequence) as $$R \to +\infty$$ exists in $$\dot{B}^{s}_{p,r}(\mathbb R^d)$$.

Finally, the limit, denoted by $$v$$, must satisfy $$\langle v, \varphi \rangle = \langle u, \varphi \rangle$$ for all test functions $$\varphi \in C^{\infty}_c(\mathbb R^d)$$. Hence, $$u$$ and $$v$$ must differ by a constant which is exactly zero.

For references on theorems used here, look at Bahouri "Fourier Analysis and Nonlinear Partial Differential Equations"