# What is difference between L2 norm and H2 Norm?

When someone refers 2-norm of system,L2 and H2 are used interchangeably by author and is rather confusing. Even the matlab has different functions for H-infinity norm and L-infinity norm.

as shown in picture the author refers it as L2, and later he refers it as H2 .

so what is difference and when someone says 2-norm of system, is he referring to L2 or H2?

I am new, so sorry for not following the norms here.

• You're more likely to get an answer to this question if you cite the source that you're referring to and quote the part of that source that refers to this as an "H2" norm. Jul 25 at 14:41

I am not sure about your application -- and we say the $$L^2$$ norm of a function and not a system. But for simplicity I will explain the concepts for real valued functions. Consider an open domain $$\Omega$$ and a function $$f:\Omega \to \mathbb{R}$$. We say that $$f \in L^2(\Omega)$$ if $$||f||_{L^2(\Omega)} < \infty$$ where $$$$||f||^2_{L^2(\Omega)} = \int_{\Omega}|f(x)|^2 dx.$$$$ Intuitively, an $$L^2$$ function is a function whose area under its graph is finite while you allow discontinuities in the function itself. For example if you take a sine wave and create a set of discontinuities of measure zero (delete "some" points on the sine wave), it will still be integrable and hence in $$L^2$$ (yet it is not continuous anymore).
Now if you consider the space $$H^1(\Omega)$$, it consists of all functions $$f$$ such that $$||f||_{H^1{\Omega}} < \infty$$ where $$$$||f||^2_{H^1(\Omega)} = \int_{\Omega}|f(x)|^2 + |f'(x)|^2 dx.$$$$ Intuitively, functions in $$H^1$$ are functions that are weakly differentiable, that is they are differentiable everywhere except at a set of points of measure 0. That means that $$f'$$ has "some discontinuity points" and so $$f' \in L^2$$. (a very nice example is the hat functions)
Finally, using the same logic, functions $$f \in H^2(\Omega)$$ are those functions that are twice - weakly differentiable and so the same logic of the previous space $$H^1$$ applies.