question about $TGV^2$ space

Question

Let us just stay in $\mathbb R^1$. The space $TGV^k$ is defined as the function $u\in L^1(I)$ and $$ TGV^k(u,I):=\sup\left\{\int_I u\,\phi^{(k)}\,d\mu, \,\phi\in C_c^\infty(I),\,\|\phi\|_{L^{\infty}(I)}\leq1,\,\|\phi'\|_{L^\infty}\leq 1,\ldots,\|\phi^{(k-1)}\|_{L^\infty}\leq 1\right\}<\infty $$ where by $\phi^{(k)}$ I mean the $k$-th derivative of $\phi$. This space suppose to generalize space $BV$ since as $k=1$ this is exactly $BV$, or $TV$, space.

Now let's assume $k=2$, i.e., we are in $TGV^2$ space. It is amazingly that we have $TGV^2$ and $BV$ is an equivalent space, i.e., $$ c\|u\|_{BV}\leq \|u\|_{L^1}+TGV^2(u)\leq C\|u\|_{BV} $$ By $\|\cdot\|_{BV}$ I mean $\|u\|_{L^1}+|\mu|_{\mathcal M}$ where $\mu$ is the measure as the weak derivative of $u$.

The prove can be found here, section 3.

It is kind of an unexpected result since with one more derivative I would expect something new. But anyway, if we accept this result, then for any $u\in TGV^2(I)$, we have $u\in BV(I)$ and there will be a Radon measure $\mu$ such that $$ \int_I u\,\varphi'dx = -\int_I \varphi\,d\mu $$ for any $\varphi\in C_c^\infty(I)$. Now if we go back to $TGV^2$, we could write $$ \int_I u\,\phi''dx = -\int_I \phi'\,d\mu $$ Then what is next? Can I write $$ \int_I u\,\phi'd\mu = -\int_I \phi\,d\nu??\tag 1 $$ for some Radon measure $\nu$? I would expect some sort of IBP formula like $$ \int_I u\,\phi''dx=\int_I \phi\,d\nu $$ to be true...

Also, the quantity $TGV^2$ I defined at the beginning, could it be explained as the total variation of a Radon measure? Like the one we used in $BV$ space? i.e., $TV(u)=|Du|$ if $u\in BV(\Omega)$. Also, some intuitive explanation of why, the $TGV^2$ norm with one more derivative, does not give any different then $BV$ norm would be really good.

Any help is really welcome!

PS: some discussion about $(1)$ can be found here.

Answer 1

My intuitive explanation that $TGV^2$ does lead to an equivalent norm on $BV$ is the following: You do not really have a higher derivative since setting $\psi = \phi^{(k-1)}$ shows that you really measure the pairing $\int u\psi' dx$ for $\|\psi\|_\infty\leq 1$. The "higher" derivatives are really lower derivatives: You only supremize the integral $\int u\psi' dx$ over some special bounded functions, namely ones that are themselves derivatives of bounded functions.

To get more intuition about the $TGV$ seminorm I suggest to look at the proof of the estimates $c\|u\|_{BV}\leq \|u\|_1 + TGV^2(u) \leq C\|u\|_{BV}$ and check how large the constants are, on what they depend (e.g. the size or shape of the domain?) and check the actual value of the norms for special cases.

There are recent papers on $TGV$ denoising in one dimension where actual minimizers are derived exactly:

• A study of the one dimensional total generalised variation regularisation problem by Konstantinos Papafitsoros and Kristian Bredies
• Exact Solutions of One-Dimensional TGV by Christiane Pöschl and Otmar Scherzer

Carefully going through the constructions and proofs should provide some further intuition.

The $TV$ seminorm is not a Radon measure. But it is the variational norm of a Radon measure, namely the distributional gradient of $u$ which is then (if $u\in BV$) a vector valued Radon measure, denoted by $Du$. Then it holds $TV(u) = |Du|(\Omega)$ where the right hand side means: Take the distributional derivative of $u$, interpret it as a vector valued Radon measure $Du$, then calculate its variation measure $|Du|$ and measure the whole set $\Omega$ with this very (now real valued) measure $|Du|$.

Since $\|u\|_1 + TGV^2(u)$ is an equivalent norm on $BV$ it follows that the if $TGV^2(u)<\infty$, then $u\in BV$ and hence, $Du$ is again a vector valued Radon measure and it pairs with continuous functions as $$\int \phi \mathrm{d} Du = \langle Du,\phi\rangle = -\langle u,\phi'\rangle = -\int u\phi'$$ as before. So $TGV^2(u) = \sup \int \phi^{(k-1)}\mathrm{d}Du$ where the supremum is taken over function $\phi$ with $\|\phi\|_\infty\leq 1$,...,$\|\phi^{(k-1)}\|_\infty\leq 1$. I don't know if you can get more explicit than that.