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I'd like to have some kind of feedback about whether or not I correctly understand the concept and if the definition I use is sensible from a formalistic mathematical standpoint (is it welldefined, does it have inconsistencies etc.).

Let $\Omega \subset R^d$, $u:\Omega \rightarrow R$ be a differentiable function and $H$ be a discrete, (equidistant) lattice on $\Omega$ with nodes $h$.

Further let $\nabla_h u$ be the gradientfield of $u$, evaluated on the nodes $h$.

Now we can define the following iterative Norm:

$$ \|\|\nabla_h u\|_p\|_q: R^d\rightarrow R, x \mapsto \sum\limits_{h\in \underline{H}}\sqrt[q]{\|\nabla_h u(h)\|_p^q}. $$

Note that the absolute value for the outer norm can be omitted, as positivity is ensured by the inner one.

Now all of the ordinary vector norm properties follow directly from the inner $p$ and $q$ norms used for the definition. Furthermore, the vector-norms used can be chosen arbitrarily, the function defined above should stay a norm regardless. For the outer norm, the nodes $h\in H$ can be interpreted as a long vector on which the norm can be applied (instead of the d-dimensional Tensor the grid would ordinarily be interpreted as).

For convenience sake I'll use the notation "$\|\cdot \|_{2,1}$" instead of writing down the double $\|$ symbols.

Now having defined this norm, I'd like to compute the convex-conjugated function $\|\|\nabla_h u\|_2\|_1^*$, where $f^*$ is defined to be

$$f^*(y) = \sup\limits_{x} \left(\langle y, x\rangle -f(x)\right).$$

I ommited a little bit of theoretical background for the fenchel conjugate, if you want me to I can jott that down too, but I'd rather not.

Now I kind of know what the fenchel conjugate is going to be and I roughly know how to arrive at that result, but I struggle to write that down correctly and I don't quite know how certain steps work yet. If someone would be able to derive that and explain it to me, then I'd be really grateful.

Finding the fenchel conjugate for the euclidean norm is very straight forward and when I tried to compute the answer for this problem I tried to go the same route. First I proofed that $\|\cdot \|_{2,1}$ is indeed a norm (homogenity is sufficient for the proof).

Then I started with the case $\|z\|_{2,1}>1$. Set $x:=\frac{t\cdot z}{\|z\|_{2,1}}$ for a $t\in R$. We then get:

$$ \langle z,x\rangle -\|x\|_{2,1}=\langle z,\frac{t\cdot z}{\|z\|_{2,1}}\rangle-\|\frac{tz}{\|z\|_{2,1}}\|_{2,1}\stackrel{Homogenity}{=}t\cdot \frac{\langle z,z\rangle}{\|z\|_{2,1}}-\frac{t}{\|z\|_{2,1}}\|z\|_{2,1}=t\left(\frac{\langle z,z \rangle}{\|z\|_{2,1}}-1\right) $$

As both $\|\cdot\|_{2,1}$ and $\|\cdot\|_2$ are norms on the finite dimensional vectorspace $R^d$ there exists $C\in R$, such that $\|x\|_2\leq C\|x\|_{2,1}~\forall ~x\in R^d$.

Plugging that into the last equation from above, we get:

$$ t\left(\frac{\langle z,z \rangle}{\|z\|_{2,1}}-1\right)\leq t\left(\frac{\left(C\|z\|_{2,1}\right)^2}{\|z\|_{2,1}}-1\right)=t\left(C^2\underbrace{\|z\|_{2,1}-1}_{>0} \right) $$ But here I am stuck again, I don't quite know how to proceed. Any good ideas from anyone?

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  • $\begingroup$ see Section 2 of math.univ-toulouse.fr/~jbhu/A_note_on_the_LF_transform.pdf $\endgroup$ – LinAlg Jul 6 '18 at 13:27
  • $\begingroup$ Thanks for the referal, I took a look at the document and while I understand why you refered me, I could not make out the exact way of how to solve my problem only using the information from the book. $\endgroup$ – sblasher Jul 9 '18 at 9:14
  • $\begingroup$ could you show your attempt at writing the conjugate? $\endgroup$ – LinAlg Jul 9 '18 at 12:46
  • $\begingroup$ yep sure, I'll edit my question. $\endgroup$ – sblasher Jul 10 '18 at 10:06
  • $\begingroup$ Hm, I just considered equality of norms on finite dimensional spaces. As $\|\cdot \|_2$ and $\|\cdot \|_{2,1}$ are defined on finite dimensional vectorspaces, there should exist positve $C,D \in R$, s.t. $C\|x\|_2\leq D\|x\|_{2,1}$ for all $x \in R^d$. Then we could plug in $\|x\|_{2,1}\geq \frac{C}{D}\|x\|_2$ into the last equation and the proof should work out exactly as in the euclidean case. $\endgroup$ – sblasher Jul 10 '18 at 10:52

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