The classical [Sobolev space](/page/Sobolev%20Space) $W^{k,2}(\mathbb{R}^n)$ consists of $L^2$ [functions](/page/Function) whose [distributional](/page/Distribution) [derivatives](/page/Derivative) up to order $k$ are also in $L^2$. This definition is limited to integer orders: the notion of "$s$ derivatives" for a non-integer $s$ — say, "$1/2$ of a derivative" — has no direct meaning in the distributional framework. Yet non-integer regularity arises naturally throughout PDE theory: elliptic [boundary](/page/Boundary) value problems produce solutions with fractional Sobolev regularity, trace operators map $H^1(\Omega)$ to $H^{1/2}(\partial\Omega)$, and interpolation between integer-order spaces lands at fractional orders. We need a definition that extends the Sobolev scale to all real $s \in \mathbb{R}$.

[motivation] ## Motivation ### The Fourier Characterisation of Integer Regularity The key insight is that the [Fourier transform](/page/Fourier%20Transform) converts the derivative count into a polynomial weight. For a [Schwartz function](/page/Schwartz%20Space) $\phi \in \mathcal{S}(\mathbb{R}^n)$, [Plancherel's theorem](/theorems/247) gives \begin{align*} \|\partial^\alpha \phi\|_{L^2}^2 &= \|\xi^\alpha \hat{\phi}\|_{L^2}^2 = \int_{\mathbb{R}^n} |\xi^\alpha|^2 \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi), \end{align*} so controlling derivatives in $L^2$ is the same as controlling polynomial moments of the Fourier transform. Summing over all multi-indices $|\alpha| \le k$: \begin{align*} \sum_{|\alpha| \le k} \|\partial^\alpha \phi\|_{L^2}^2 &= \int_{\mathbb{R}^n} \sum_{|\alpha| \le k} |\xi^\alpha|^2 \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi) \sim \int_{\mathbb{R}^n} (1+|\xi|^2)^k \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi), \end{align*} where $\sim$ denotes two-sided bounds with constants depending only on $n$ and $k$. **Having $k$ derivatives in $L^2$ is equivalent to the Fourier transform being square-[integrable](/page/Integral) against the weight $(1+|\xi|^2)^k$.** ### From Integers to All Reals The weight $(1+|\xi|^2)^k$ makes sense for any real exponent, not just integers. Replacing $k$ by $s \in \mathbb{R}$ gives a meaningful condition \begin{align*} \int_{\mathbb{R}^n} (1+|\xi|^2)^s \, |\hat{\phi}(\xi)|^2 \, d\mathcal{L}^n(\xi) &< \infty \end{align*} that interpolates continuously between integer regularities. The frequency variable $\xi$ encodes spatial oscillation: large $|\xi|$ corresponds to rapid oscillation (high frequency), small $|\xi|$ to slow variation (low frequency). For $s > 0$, the weight penalises high-frequency components, enforcing smoothness. For $s < 0$, the weight *amplifies* high frequencies, allowing objects rougher than $L^2$ — such as the Dirac delta, which belongs to $H^s(\mathbb{R}^n)$ precisely when $s < -n/2$ (Problem 2). The "$+1$" in $(1+|\xi|^2)$ ensures that low-frequency components are also controlled, giving $L^2$ membership as a baseline — this is what distinguishes the inhomogeneous spaces from their [homogeneous counterparts](/page/Homogeneous%20Sobolev%20Space). ### The Distributional Subtlety For a Schwartz function $\phi$, the integral above involves $|\hat{\phi}(\xi)|^2$ — the pointwise square of a genuine function. But for a [tempered distribution](/page/Tempered%20Distributions) $u \in \mathcal{S}'(\mathbb{R}^n)$, the Fourier transform $\hat{u}$ is another tempered distribution, not a function, and "$|\hat{u}(\xi)|^2$" has no meaning. The correct definition uses the distributional product $(1+|\xi|^2)^{s/2}\hat{u}$ — the product of the slowly increasing function $(1+|\xi|^2)^{s/2} \in \mathcal{O}_M(\mathbb{R}^n)$ with the tempered distribution $\hat{u}$, as defined on the [Tempered Distributions](/page/Tempered%20Distributions) page — and asks whether this tempered distribution is represented by an $L^2$ function. [/motivation]

## Definition

[definition: Inhomogeneous Sobolev Space] Let $s \in \mathbb{R}$ and $n \ge 1$. The **inhomogeneous Sobolev space** of order $s$ is \begin{align*} H^s(\mathbb{R}^n) &:= \left\{u \in \mathcal{S}'(\mathbb{R}^n) \;\middle|\; \text{there exists } g \in L^2(\mathbb{R}^n) \text{ such that } (1+|\xi|^2)^{s/2}\hat{u} = T_g\right\}, \end{align*} where $T_g: \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}$ is the [regular distribution](/page/Regular%20Distribution) associated to $g$, defined by \begin{align*} T_g(\phi) &:= \int_{\mathbb{R}^n} g(\xi)\,\phi(\xi) \, d\mathcal{L}^n(\xi), \end{align*} and $(1+|\xi|^2)^{s/2}\hat{u}$ denotes the product of the slowly increasing function $(1+|\xi|^2)^{s/2} \in \mathcal{O}_M(\mathbb{R}^n)$ with the tempered distribution $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$, acting on [test functions](/page/Test%20Function) by \begin{align*} \bigl((1+|\xi|^2)^{s/2}\hat{u}\bigr)(\phi) &= \hat{u}\bigl((1+|\xi|^2)^{s/2}\phi\bigr) \quad \text{for every } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} The **norm** is $\|u\|_{H^s} := \|g\|_{L^2(\mathbb{R}^n)}$. [/definition]

Three things must be verified: that the product $(1+|\xi|^2)^{s/2}\hat{u}$ is well-defined for any $u \in \mathcal{S}'$, that the $L^2$ representative $g$ is unique ($\mathcal{L}^n$-a.e.), and that the resulting normed space is complete. The following theorem establishes all three by constructing an isometric isomorphism $\Phi_s: H^s \to L^2$ via $\Phi_s(u) = g$.

[quotetheorem:466]

The map $\Phi_s$ "flattens" regularity by reweighting in frequency space, reducing every element of $H^s$ to a plain $L^2$ function. Since $\Phi_s$ is an isometric bijection onto $L^2$, the space $H^s$ inherits all the structure of $L^2$: it is a [separable](/page/Separable) [Hilbert space](/page/Hilbert%20Space) with inner product

\begin{align*} \langle u, v \rangle_{H^s} &= \langle \Phi_s(u), \Phi_s(v) \rangle_{L^2} = \int_{\mathbb{R}^n} g_u(\xi)\,\overline{g_v(\xi)} \, d\mathcal{L}^n(\xi), \end{align*}

where $g_u, g_v \in L^2(\mathbb{R}^n)$ represent $(1+|\xi|^2)^{s/2}\hat{u}$ and $(1+|\xi|^2)^{s/2}\hat{v}$.

### The Integral Formula

Once membership $u \in H^s$ is established, the Fourier transform $\hat{u}$ turns out to be more than just a tempered distribution — it is represented by a locally integrable function, and the $H^s$ norm can be written as a genuine integral involving this function. The following result makes this precise.

[quotetheorem:929]

The representative $(\hat{u})_{\mathrm{rep}}$ is related to the $L^2$ function $g = \Phi_s(u)$ from [Theorem 466](/theorems/466) by $(\hat{u})_{\mathrm{rep}}(\xi) = g(\xi)(1 + |\xi|^2)^{-s/2}$, $\mathcal{L}^n$-a.e. The proof proceeds by dividing the membership condition $(1 + |\xi|^2)^{s/2}\hat{u} = T_g$ by the smooth, positive weight $(1 + |\xi|^2)^{s/2} \in \mathcal{O}_M(\mathbb{R}^n)$, which preserves regularity of the distribution. The full proof is available on the [theorem page](/theorems/929).

The integral formula is the version appearing in most textbooks as $\int (1+|\xi|^2)^s|\hat{u}(\xi)|^2\,d\mathcal{L}^n$, with "$\hat{u}(\xi)$" silently denoting $(\hat{u})_{\mathrm{rep}}(\xi)$.

[remark: The Integral Formula for $L^2$ Functions] When $f \in L^2(\mathbb{R}^n)$, the [recovery theorem](/theorems/928) gives $\iota(f) = T_f \in H^s(\mathbb{R}^n)$ whenever $(1 + |\xi|^2)^{s/2}\hat{f} \in L^2(\mathbb{R}^n)$. By the [Plancherel theorem](/theorems/247), $\widehat{T_f} = T_{\hat{f}}$ where $\hat{f} \in L^2(\mathbb{R}^n)$ is the $L^2$ Fourier transform of $f$. The representative is then $(\widehat{T_f})_{\mathrm{rep}} = \hat{f}$, and the integral formula becomes \begin{align*} \|\iota(f)\|_{H^s}^2 = \int_{\mathbb{R}^n} (1 + |\xi|^2)^{s}\, |\hat{f}(\xi)|^2\, d\mathcal{L}^n(\xi), \end{align*} where $\hat{f}$ is a genuine $L^2$ function — no distributional products or $T$-representatives are involved. This is the formula used throughout PDE theory whenever the function in question is known to be in $L^2$. [/remark]

## Recovery of Classical Sobolev Spaces

The Fourier-based definition of $H^s$ must agree with the classical [Sobolev space](/page/Sobolev%20Space) $W^{k,2}$ at integer orders. However, the objects are of different types: elements of $W^{k,2}(\mathbb{R}^n)$ are (equivalence classes of) functions, while elements of $H^k(\mathbb{R}^n)$ are tempered distributions. The correct statement is not that the spaces are "equal," but that the canonical embedding $f \mapsto T_f$ — sending a function to its regular distribution — is an isomorphism with equivalent norms.

[quotetheorem:928]

The distinction between "equivalent" and "equal" norms is genuine: the $H^k$ norm uses the weight $(1 + |\xi|^2)^k$, which is a single monomial in $|\xi|^2$, while the $W^{k,2}$ norm uses the sum $\sum_{|\alpha| \le k} |\xi^\alpha|^2$, a sum of many monomials. These are comparable (both are degree-$k$ polynomials in $|\xi|$ that agree at leading order) but not identical — the equivalence constants depend on the number of multi-indices, which grows with $n$ and $k$. The special case $k = 0$ is the exception: both weights equal $1$, so the isomorphism $L^2 \cong H^0$ is isometric.

In practice, when $f \in W^{k,2}(\mathbb{R}^n)$ and the context involves $H^k$, one must apply the embedding $\iota$ explicitly: the correct statement is $\iota(f) = T_f \in H^k(\mathbb{R}^n)$, not "$f \in H^k$." The norm comparison then reads $\|\iota(f)\|_{H^k} \sim \|f\|_{W^{k,2}}$, with the map $\iota$ mediating between the two spaces.