Definition of Sobolev space $H^s$ and domain of $-\Delta^s$ The spaces below are on $\partial\Omega$, the boundary of a bounded smooth domain $\Omega$. I read this in the book on page 141.


  Define $H^2 := \{ u \in L^2 \mid (-\Delta u) \in L^2\}$.


And then:


  We can define $H^s$ ($s \in \mathbb{R}$) as the domain of $(-\Delta)^s$. If $\lambda_j, e_j$ are the eigenpairs of $-\Delta$, then
  $$H^s = \{ u \in (C_0^\infty)' \mid  \sum_j\lambda_j^{2s}|(u,e_j)_{L^2}|^2 &lt; \infty \}$$


Now I'm confused, shouldn't the power of the eigenvalues be $s$, not $2s$? 

For example take $u \in H^1$. Then $u=\sum (u,e_j)e_j$, so $-\Delta u = \sum \lambda_j(u,e_j)e_j$ in the weak sense, so $\langle -\Delta u, u \rangle = \sum \lambda_j |(u,e_j)|^2 &lt; \infty$. But with the definition of $H^1$ above, we would need $\sum \lambda_j^2|(u,e_j)|^2 &lt; \infty$. Why this discrepancy?
You're looking at the square of the norm.  In the last line of example 4.17, note the square root of the sum of squares.

It's natural to see the squares of the eigenvalues in the square of the norm.  Both the norm and the square are going to be finite, and can be used to define $H^s$, but the square isn't going to be homogeneous (like you're expecting).

EDIT: again, what you're looking at is the square of a norm.  In the example you cite, $\langle -\Delta u, u \rangle$ is the square of a norm on $u$ in $H^{1/2}$ on the boundary.

Consider an analogy in $R^n$ with a symmetric,   positive-definite matrix $A$ with the $A$-norm $\left\|x\right\|_A^2 = x^TAx = \sum \lambda_i (x\cdot u_i)^2$ where the $\lambda_i$ are the (positive) eigenvalues of $A$ and $u_i$ the corresponding eigenvectors.  Note that $\lambda_i=\left(\lambda_i^{1/2}\right)^2$.  Another way to rewrite this is $\left\|x\right\|_A = \left\|A^{1/2}x\right\|$.

EDIT2: Or is the confusion from the interplay between the interior and the boundary?  $\langle -\Delta u, u \rangle$ isn't the square of a norm on $H^1$ on the boundary (but is on the interior).

EDIT3: There is a bounded, linear operator -- the trace operator -- that maps from $H^1(\Omega)$ to $H^{1/2}(\partial\Omega)$.  It is the operator that finds an extension of a function defined on the interior of a domain to its boundary.  The integration by parts formula you cite is part and parcel of the proof that such a map exists.

Note that a function $u\in H^1(\Omega)$ is not guaranteed to have an extension $u|_{\partial\Omega}$ in $H^1(\partial\Omega)$.  That is, the values of the extension on the boundary are slightly less regular than the values of the function on the interior.  In the other direction, the interpolation construction in Oden's book is necessary to show that the extension is in $H^{1/2}(\partial\Omega)$ and not just $L^2(\partial\Omega)$.
You're looking at the square of the norm.  In the last line of example 4.17, note the square root of the sum of squares.

It's natural to see the squares of the eigenvalues in the square of the norm.  Both the norm and the square are going to be finite, and can be used to define $H^s$, but the square isn't going to be homogeneous (like you're expecting).

EDIT: again, what you're looking at is the square of a norm.  In the example you cite, $\langle -\Delta u, u \rangle$ is the square of a norm on $u$ in $H^{1/2}$ on the boundary.

Consider an analogy in $R^n$ with a symmetric,   positive-definite matrix $A$ with the $A$-norm $\left\|x\right\|_A^2 = x^TAx = \sum \lambda_i (x\cdot u_i)^2$ where the $\lambda_i$ are the (positive) eigenvalues of $A$ and $u_i$ the corresponding eigenvectors.  Note that $\lambda_i=\left(\lambda_i^{1/2}\right)^2$.  Another way to rewrite this is $\left\|x\right\|_A = \left\|A^{1/2}x\right\|$.

EDIT2: Or is the confusion from the interplay between the interior and the boundary?  $\langle -\Delta u, u \rangle$ isn't the square of a norm on $H^1$ on the boundary (but is on the interior).

EDIT3: There is a bounded, linear operator -- the trace operator -- that maps from $H^1(\Omega)$ to $H^{1/2}(\partial\Omega)$.  It is the operator that finds an extension of a function defined on the interior of a domain to its boundary.  The integration by parts formula you cite is part and parcel of the proof that such a map exists.

Note that a function $u\in H^1(\Omega)$ is not guaranteed to have an extension $u|_{\partial\Omega}$ in $H^1(\partial\Omega)$.  That is, the values of the extension on the boundary are slightly less regular than the values of the function on the interior.  In the other direction, the interpolation construction in Oden's book is necessary to show that the extension is in $H^{1/2}(\partial\Omega)$ and not just $L^2(\partial\Omega)$.