# Justification for Squared ${L}_{2}$ Data and Smoothness Term as an Error Bound

Often in variational methods (and not only) we have an energy that is of the form: $$E(u) = \frac{1}{2}\|f-u\|^2_2 + \frac{\alpha}{2}\|\psi(u)\|^2_2,$$

where the first term is referred to as the data term, and the second as the smoothness term. I understand that the squared $$L_2$$ norm is especially appealing since it results in a simple expression when we are looking for the minimum. I am looking for the formal motivation for designing such an energy. More specifically, I would like to see it as resulting from some bound upon the actual error. Let $$f$$ be a degraded/noisy image of the original image $$g$$, and $$u$$ be the image that we are looking for. Is it possible for the above to be interpreted as a bound of $$\|g-f\|^2_2$$? Let us consider an even simpler problem (where $$h$$ is a sufficiently smoothed version of $$f$$): $$E(u) = \frac{1}{2}\|f-u\|^2_2 + \frac{\alpha}{2}\|u-h\|^2_2$$ If I start from $$\|g-f\|_2$$ I can get the following bound: $$\|u-g\| = \|u-f+f-g\| \leq \|u-f\| + \|f-g\|$$ $$\|u-g\| = \|u-h+h-g\| \leq \|u-h\| + \|h-g\|$$ Combining the two inequalities with some weight $$\lambda \in [0,1]$$ I get: $$\|u-g\|\leq (1-\lambda)\|u-f\| + \lambda\|u-h\| + (1-\lambda)\|f-g\| + \lambda\|h-g\|$$ Since $$f,g,h$$ are fixed, the minimization is only over the first two terms: $$\min_u (1-\lambda)\|u-f\| + \lambda\|u-h\|$$ And this can be rewritten through $$\alpha$$ by setting $$\alpha = \frac{\lambda}{1-\lambda}$$. My issue is that these are still not squared norms. If I try to square both sides, this results in a different minimization energy. The other way to get the squares involves writing out something of the form: $$\|u-g\|^2 = \|(u-f) + (f-g)\|^2 = \|u-f\|^2 + \|f-g\|^2 - 2(u-f)\cdot(g-f) \\ \leq \|u-f\|^2 + \|f-g\|^2$$ The last inequality holds only when $$(u-f)\cdot(g-f) \geq 0$$ though. Said otherwise, the angle between $$(u-f)$$ and $$(g-f)$$ needs to be smaller than 90 degrees which is not something that can be guaranteed I believe. A looser bound is: $$\|u-g\|^2 \leq \|u-f\|^2 + \|f-g\|^2 + 2\|u-f\|\|f-g\|,$$ but it also involves non-squared terms.

Here's something else I noticed about the non-squared energy:

$$\alpha\|u-f\| + \beta\|u-h\|, \alpha, \beta > 0$$

will always have as solutions $$u=f$$ or $$u=h$$ for $$\alpha<\beta$$ respectively $$\alpha>\beta$$. When $$\alpha=\beta$$ any solution of the form $$(1-\lambda)f + \lambda h$$ is admissible. It is thus clear that the above is not very interesting as a minimization energy.

Cross-posted to: https://math.stackexchange.com/q/3692290/463794 I expect to get a different perspective there.

One of the motivations to use the $${L}_{2}$$ norm comes from the Maximum a Posteriori Estimation (MAP) framework.
If you model $$\psi \left( u \right) \sim \mathcal{N} \left( 0, \alpha \right)$$ then if you derive the MAP Estimator in case the added noise is Gaussian you'd get the exact model you posted above.
• Are you aware of any different interpretations? Or maybe a resource on the topic I could refer to? In general I do not want to assume Gaussian noise, and would prefer if I could produce the quadratic energy from a bound. In that regard I guess I could use the looser bound that I have above and drop terms that involve $g$ because it is unknown, but that seems like a rather handwavy justification. – lightxbulb May 27 '20 at 14:05