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.