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Consider real variables $x, y$ and a function $f(x, y) \in H^s(\mathbb{R}^2)$, say for some $s \in (0, 1)$. I am trying to get an understanding of mixed Sobolev spaces of the form $H^s_x(H^s_y)$, where the mixed Sobolev norm is defined by $$ \Vert f\Vert_{H^s_x(H^s_y)} = \Vert \Vert f\Vert_{H^s_y}\Vert_{H^s_x}.$$ In other words, $H^s_x$ functions taking values in $H^s_y$. My question is, does $f \in H^s(\mathbb{R}^2)$ automatically imply $f \in H^s_x(H^s_y)$?

It is obvious that $L^2_x(L^2_y) = L^2(\mathbb{R}^2)$, but in the case of Sobolev spaces, it is not so clear. Understanding this (relatively simple ?) case will help me understand the general theory better.

Edit: From Lions and Magenes, and also from Michael Renardy's answer below, I think that the correct thing to try to infer is $f \in H^s(\mathbb{R}^2) \Rightarrow f \in H^s_x(L^2_y)$. But it is not clear to me how to prove this.

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    $\begingroup$ These are called anisotropic Sobolev spaces. You can find a nice presentation in Chapter 4 (vol 2) of Lions & Magenes book. $\endgroup$
    – Liviu Nicolaescu
    Commented May 10, 2015 at 14:46
  • $\begingroup$ @LiviuNicolaescu Thanks for the comment. I am looking up Lions and Magenes and have edited the question accordingly. $\endgroup$
    – anonymous
    Commented May 10, 2015 at 16:34

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The answer to your question is no. If a function $f$ is in $H^1(R^2)$, it means that $f_x$ and $f_y$ are in $L^2$, but if $f$ is in $H^1(H^1)$, it means that $f_{xy}$ is in $L^2$.

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  • $\begingroup$ I looked at Lions and Magenes, and I think that the correct inference should be $f \in H^s_x(L^2_y)$. Is this correct? I am running into trouble trying to prove this directly from the definition. Could you please give me a hint how to prove this? $\endgroup$
    – anonymous
    Commented May 10, 2015 at 17:16
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    $\begingroup$ The correct inference is that $f\in H^s(L^2)\cap L^2(H^s)$. You should look at the equivalent statement for Fourier transforms. $\endgroup$
    – Michael Renardy
    Commented May 10, 2015 at 19:18
  • $\begingroup$ I think: $f \in H^s_x(L^2_y) \Leftrightarrow \Vert (1 + \xi^2)^{s/2}\hat{f}^x(\xi)\Vert_{L^2_y} \in L^2_\xi$, where $\hat{f}^x$ means Fourier transform with respect to the variable $x$ only. This holds iff $\Vert \Vert (1 + \xi^2)^{s/2}\hat{f}^x(\xi)\Vert_{L^2_y} \Vert_{L^2_\xi} < \infty$, which would follow from $f \in H^s$. I would really appreciate your opinion on whether I understand this correctly. $\endgroup$
    – anonymous
    Commented May 10, 2015 at 20:21
  • $\begingroup$ The other inference $f \in L^2_x(H^s_y)$ would also follow similarly. $\endgroup$
    – anonymous
    Commented May 10, 2015 at 20:23
  • $\begingroup$ @MichaelRenardy I read somewhere that the correct interpretation might be $f \in H^{s/2}_x(H^{s/2}_y)$ by interpolation. If this is right, could you point to a reference? Is this interpolation result in Lions and Magenes? $\endgroup$
    – Peter Halburt
    Commented Jun 9, 2015 at 1:13

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