# High order BV function has continuous representative

As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, ${W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}$.

Definition A function ${f \in BV_n({\mathbb R}^n)}$, if ${f \in W^{n-1,1}({\mathbb R}^n)}$, and the nth order distributional derivative ${D^n f}$ is a finite Radon measure.

Theorem If ${f \in BV_n({\mathbb R}^n)}$, then ${f}$ has a continuous representative.

Proof: Note that ${C_c^\infty({\mathbb R}^n)}$ is dense in ${BV_n({\mathbb R}^n)}$, i.e. for any ${f \in BV_n({\mathbb R}^n)}$, there exists a sequence ${f_k \in C_c^\infty({\mathbb R}^n)}$ such that ${\lim_{k \rightarrow \infty} f_k = f}$ in ${W^{n-1,1}}$, and ${\lim_{k\rightarrow \infty} \| \nabla^n f_k\|_{L^1({\mathbb R}^n)} = \|D^n f \| ({\mathbb R}^n)}$. (See for example, book by Evans and Gariepy). So it suffice to show for all ${f \in C_c^\infty({\mathbb R}^n)}$,

$\displaystyle \|f\|_{L^\infty({\mathbb R}^n)} \le \| \nabla^n f\|_{L^1({\mathbb R}^n)}$

Indeed,

$\displaystyle f(x_1, \cdots , x_n) = \int_{-\infty}^{x_1} \partial_1 f(s_1,x_2, \cdots , x_n)\,ds_1 = \int_{-\infty}^{x_1} \cdots \int_{-\infty}^{x_n} \partial_1 \cdots \partial_n f$

This gives us the desire inequality. Then by density result we know that ${BV_n({\mathbb R}^n)}$ admits continuous representative for every element. $\Box$

Corollary ${W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}$.

Remark If the dimension is greater than the order of BV function, then the above theorem will fail. For example, let ${f(x) = |x|^{-\frac{1}{2}}}$ be defined in a neighborhood ${\Omega}$ around 0, with smooth boundary in ${{\mathbb R}^3}$. ${|\nabla f| \sim |x|^{-\frac{3}{2}}}$ and ${|\nabla^2 f| \sim |x|^{-\frac{5}{2}}}$. So ${f \in W^{2,1}(\Omega)}$. Then we can extend ${f}$ on ${{\mathbb R}^3}$, since ${\partial \Omega}$ is smooth (See for example, PDE book by Evans). But then ${f \rightarrow \infty}$ as ${|x| \rightarrow 0}$, which doesn’t admit a continuous representative.

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