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 fig.

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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jul 25, 2021 at 14:41

1 Answer 1


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 \begin{equation} ||f||^2_{L^2(\Omega)} = \int_{\Omega}|f(x)|^2 dx. \end{equation} 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 \begin{equation} ||f||^2_{H^1(\Omega)} = \int_{\Omega}|f(x)|^2 + |f'(x)|^2 dx. \end{equation} 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.

  • 1
    $\begingroup$ I am afraid that this answer is missing some important context. In control theory there is a standard definition of norms for control systems that does not coincide with the definition of matrix or function norms; see e.g. Section 2.2 in users.abo.fi/htoivone/courses/robust/rob2.pdf . Clearly OP is confused about those. $\endgroup$ Jul 27, 2021 at 6:23
  • $\begingroup$ @FedericoPoloni I agree ... but that is why I started by mentioning that "I do not know your application" and then gave a rather general and pure mathematical definitions and intuitions :) (the tags were added after my post) $\endgroup$
    – outlaw
    Jul 27, 2021 at 6:28
  • $\begingroup$ I understand --- I was just pointing that out for future reference and for OP. $\endgroup$ Jul 27, 2021 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.