I would like a general explanation on the difference between homogenous and inhomogeneous function spaces, there doesn't seem to be a very good explanation online.

I know that for Sobolev spaces for periodic functions, the difference between the norms for inhomogeneous $H^s$ and homogeneous $\dot{H}^s$ spaces is in modifying the weighting in the norm from $\langle k \rangle = (1 + k^2)^{1/2}$ to $|k|$. If $u$ is a $2\pi$-periodic function, it has a Fourier series representation $$u = \sum_{k \in \mathbb{Z}} u_k e^{ikx}$$ Then $$\| u \|_{H^s}^2 = \sum_{k \in \mathbb{Z}} \langle k \rangle^{2s} |u_k|^2 $$ and $$\| u \|_{H^s}^2 = \sum_{k \in \mathbb{Z}} |k|^{2s} |u_k|^2 $$ It is easy to show that $\dot{H}^s$ is the subspace of $H^s$ with zero spatial average over one period, in terms of Fourier series, the zero mode is zero.

However I am confused about the difference between inhomogeneous and homogeneous versions of other more complicated function spaces, in particular Besov spaces. The definitions in the literature are so widely varied that I can't nail down what characterises the homogeneous Besov space.

Thanks :)

  • $\begingroup$ Hi, do you have a reference on the representation of Sobolev spaces in Fourier series? I am particularly interested in the two norms defined above. $\endgroup$
    – newbie
    Commented Jan 16, 2017 at 14:59

1 Answer 1


It's not entirely clear to me whether you're asking about the definition of homogeneous and inhomogeneous Besov spaces or about what the difference in the definitions really means. Regarding both questions:

I think the answer will work better on the line instead of the circle. Let's say a norm is homogeneous if there exists an exponent $\gamma$ such that $$g(t)=f(\delta t)$$implies $$||g||=\delta^\gamma ||f||.$$ Making up a term, say a norm is ``essentially homogeneous'' if there exist $A$ and $B$ so $$A||g|\le\delta^\gamma||f||\le B||g||$$(independent of $\delta$).

Now the homogeneous Sobolev norm $$||f||^2=\int_{\Bbb R}|\xi|^{2\alpha}|\hat f(\xi)|^2\,d\xi$$is in fact a homogeneous norm in this sense, while the inhomogeneous Sobolev norm $$||f||^2=\int_{\Bbb R}(1+|\xi|^2)^{\alpha}|\hat f(\xi)|^2\,d\xi$$is not (it's not essentially homogeneous either).

Similarly for Besov spaces. There are many equivalent norms out there; the one I often find easiest to use is the one that forced me to invent the term ``essentially homogeneous''. Roughly speaking, say $$f=F+\sum_{1}^\infty f_n=\sum_{-\infty}^\infty f_n,$$where $\hat F$ is supported in the interval $[-2,2]$ and $\hat f_n$ is supported in the annulus $2^{n-1}<|\xi|<2^{n+1}$. The inhomogeneous Besov space norm $$||F||_p+\left(\sum_{n=1}^\infty(2^{n\alpha}||f_n||_p)^q\right)^{1/q}$$is not even essentially homogeneous, while the homogeneous Besov space norm $$\left(\sum_{n=-\infty}^\infty(2^{n\alpha}||f_n||_p)^q\right)^{1/q}$$_is_ essentially homogeneous (it's precisely homogeneous for $\delta=2^n$, while the norm of $g$ for $2^n<\delta<2^{n+1}$ is essentially the same as for $\delta=2^n$). Some other versions of the Besov space norm are in fact exactly homogenous in the homogeneous case.

Edit, having had the question clarified somewhat:

In what seems to me the typical case the norm on the inhomogeneous space is the sum of a norm plus a homogeneous seminorm. For example, a norm equivalent to the inhomogeneous Sobolev norm above (equivalent when $\alpha\ge0$) is $$||f||_2+\left(\int|\xi|^{2\alpha}|\hat f(\xi|)|^2\right)^{1/2},$$ while a norm equivalent to the inhomogeneous Besov space norm above is $$||f||_p+\left(\sum_{n=-\infty}^\infty(2^{n\alpha}||f_n||_p)^q\right)^{1/q}.$$The versions II gae previously are perhaps easier to deal with and more standard, but these versions are maybe better for the current discussion because you can see that the norm is the sum of two parts; each part is homogeneous, but of different degress, so the sum becomes inhomogeneous.

I think it may be fair to say that the homogeneous norm is nice because of the homogeneity. But it's not nice because it's not actually a norm, only a seminnorm; roughly speaking, in these situations the homogeneous norm measures the size of the derivatives of $f$, not $f$ itself. So a function in the homogeneous Sobolev or Besov space need not even be in the relevant $L^p$ space. We add $||f||_p$ to the homogeneous norm and now we have an actual norm, defined for actual $L^p$ functions. But we lose homogeneity.

  • $\begingroup$ Hi David. Thanks for your answer. I understand my question was a bit confusingly worded. What I was trying to get at was when people talk about some space $X$ and the homogeneous subspace $\dot{X}$, what is the general rule for what is in $\dot{X}$. I understand that for function spaces, we usually take the homogeneous subspace as functions with zero average/Fourier transforms not having support at the origin, and this results in a simpler norm with a homogeneity property. Is that how we decide what the homogeneous subspace is? Just see what simplification makes the norm have this property!?!? $\endgroup$
    – Frubiclé
    Commented Mar 5, 2016 at 0:01
  • 2
    $\begingroup$ In fact typically the homogeneous space is not a subspace of the inhomogeneous space, it's the other way around. Probably you've been fooled by concentrating on the circle - in fact the homogenneous and inhomgeneous spaces on the circle are the same space, with different norms. The rest of your question is maybe answered in an edit to my answer. $\endgroup$ Commented Mar 5, 2016 at 0:47
  • $\begingroup$ Erm, regarding what I said about the spaces on the circle being the same: That's the way it would be if we directly translate the definitions from the line to the circle. Come to think of it, then people fix the fact that it's just a seminorm by excluding the constants, requiring integral zero. And then sure enough the homogeneous space is a subspace, as you said. I think that this is really hiding the essential points - the whole notion of "homogeneity" makes less sens on the circle, since there are no dilations. Imo what it all really means is as I said on the line. $\endgroup$ Commented Mar 5, 2016 at 0:56

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