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. [example: $H^1$ Norm Of A Gaussian] Let $n = 1$ and $u(x) = e^{-x^2/2}$, a Gaussian. Its Fourier transform is $\hat{u}(\xi) = \sqrt{2\pi}\,e^{-2\pi^2\xi^2}$ (up to convention). The $H^1$ norm is \begin{align*} \|u\|_{H^1}^2 &= \int_{\mathbb{R}} (1+\xi^2) \, |\hat{u}(\xi)|^2 \, d\mathcal{L}^1(\xi) = 2\pi \int_{\mathbb{R}} (1+\xi^2)e^{-4\pi^2\xi^2}\,d\mathcal{L}^1(\xi). \end{align*} The integral splits into two Gaussian moments: \begin{align*} \int_{\mathbb{R}} e^{-4\pi^2\xi^2}\,d\mathcal{L}^1(\xi) &= \frac{1}{2\pi}, \quad \int_{\mathbb{R}} \xi^2 e^{-4\pi^2\xi^2}\,d\mathcal{L}^1(\xi) = \frac{1}{16\pi^3}. \end{align*} Therefore $\|u\|_{H^1}^2 = 2\pi\bigl(\frac{1}{2\pi} + \frac{1}{16\pi^3}\bigr) = 1 + \frac{1}{8\pi^2}$, which matches the classical computation $\|u\|_{L^2}^2 + \|u'\|_{L^2}^2$ (since $u'(x) = -xe^{-x^2/2}$). [/example] ### Negative Orders and Duality For negative integer $s = -k$, the space $H^{-k}(\mathbb{R}^n)$ consists of tempered distributions expressible as finite sums of distributional derivatives of $L^2$ functions up to order $k$: \begin{align*} u \in H^{-k}(\mathbb{R}^n) \quad &\Longleftrightarrow \quad u = \sum_{|\alpha| \le k} \partial^\alpha f_\alpha \text{ for some } f_\alpha \in L^2(\mathbb{R}^n). \end{align*} The space $H^{-k}$ is the dual of $H^k$ under the $L^2$ pairing: every [continuous](/page/Continuity) linear functional on $H^k(\mathbb{R}^n)$ is represented by a unique element of $H^{-k}(\mathbb{R}^n)$, and the operator norm equals the $H^{-k}$ norm. The isometry $\Phi_s$ makes this transparent: the dual of $H^k$ is $(H^k)^* \cong L^2 \cong H^{-k}$ via the chain $\Phi_k^* \circ \Phi_{-k}^{-1}$. ## The Sobolev Embedding Threshold The most structurally significant property of the $H^s$ scale is that membership in $H^s$ for $s > n/2$ forces every element to have a continuous representative. The mechanism is that $s > n/2$ makes the Fourier transform absolutely integrable, allowing Fourier inversion to produce a pointwise-defined continuous function. This threshold is *sharp*: for $s \le n/2$, there exist elements of $H^s$ that are discontinuous or unbounded. [quotetheorem:226] The proof hinges on a single application of Cauchy–Schwarz: inserting the weight $(1+|\xi|^2)^{s/2}$ and its reciprocal, \begin{align*} \|\hat{u}_{\mathrm{rep}}\|_{L^1} &= \int_{\mathbb{R}^n} (1+|\xi|^2)^{-s/2} \cdot (1+|\xi|^2)^{s/2}|\hat{u}_{\mathrm{rep}}(\xi)|\,d\mathcal{L}^n(\xi) \\ &\le \left(\int_{\mathbb{R}^n} (1+|\xi|^2)^{-s}\,d\mathcal{L}^n(\xi)\right)^{1/2} \|u\|_{H^s}, \end{align*} and the weight integral converges if and only if $s > n/2$ (the integrand decays like $|\xi|^{-2s}$ and $\int_1^\infty r^{n-1-2s}\,dr < \infty$ iff $2s > n$). Once $\hat{u}_{\mathrm{rep}} \in L^1$, Fourier inversion gives a bounded continuous representative. When $s > n/2 + k$ for an integer $k \ge 0$, the embedding sharpens: elements of $H^s$ have $k$ continuous derivatives. This follows by applying the [Sobolev embedding](/theorems/226) to $\partial^\alpha u \in H^{s-k}$ for $|{\alpha}| \le k$, each of which has $s - k > n/2$. [example: Sharpness Of The Embedding Threshold] The indicator function $f = \mathbb{1}_{[-1,1]}$ in dimension $n = 1$ is discontinuous at $\pm 1$. Its Fourier transform decays like $|\hat{f}(\xi)| \sim |\xi|^{-1}$ for large $|\xi|$: \begin{align*} \hat{f}(\xi) &= \int_{-1}^1 e^{-2\pi i \xi x}\,d\mathcal{L}^1(x) = \frac{\sin(2\pi\xi)}{\pi\xi}. \end{align*} The $H^s$ norm finiteness reduces to the tail integral \begin{align*} \int_1^\infty (1+\xi^2)^s |\hat{f}(\xi)|^2\,d\mathcal{L}^1(\xi) &\sim \int_1^\infty \xi^{2s} \cdot \xi^{-2}\,d\mathcal{L}^1(\xi) = \int_1^\infty \xi^{2s-2}\,d\mathcal{L}^1(\xi), \end{align*} which converges if and only if $2s - 2 < -1$, i.e., $s < 1/2 = n/2$. So $f \in H^s(\mathbb{R})$ for every $s < 1/2$ but $f \notin H^{1/2}(\mathbb{R})$: the embedding threshold is exactly $n/2$. [/example] ## Elliptic Regularity on the $H^s$ Scale The defining feature of inhomogeneous Sobolev spaces in PDE theory is that elliptic operators shift regularity along the $H^s$ scale. An elliptic operator of order $m$ maps $H^{s+m} \to H^s$, and invertible elliptic operators shift in the reverse direction. This is the mechanism by which elliptic PDEs gain regularity: if the data $f$ belongs to $H^s$, the solution $u$ belongs to $H^{s+m}$ — the operator "promotes" the regularity by its order. The simplest and most important instance is the Helmholtz operator $(1-\Delta)$, whose Fourier symbol is $(1+|\xi|^2)$ — exactly the weight defining the $H^s$ norm. [quotetheorem:227] The [Helmholtz isomorphism](/theorems/227) is completely transparent on the Fourier side: the equation $(1-\Delta)u = f$ becomes $(1+|\xi|^2)\hat{u} = \hat{f}$, so \begin{align*} \hat{u}(\xi) &= \frac{\hat{f}(\xi)}{1+|\xi|^2}. \end{align*} The factor $(1+|\xi|^2)^{-1}$ suppresses high-frequency components by exactly two orders, promoting $H^s$ data to an $H^{s+2}$ solution. The isometry is equally transparent: \begin{align*} \|u\|_{H^{s+2}} &= \|(1+|\xi|^2)^{(s+2)/2}\hat{u}\|_{L^2} = \|(1+|\xi|^2)^{s/2}\hat{f}\|_{L^2} = \|f\|_{H^s}. \end{align*} [example: Regularity Gain For The Helmholtz Equation] Let $n = 3$ and $f = \Delta g$ for some $g \in H^1(\mathbb{R}^3)$, so $f \in H^{-1}(\mathbb{R}^3)$. The solution $u = (1-\Delta)^{-1}f \in H^1(\mathbb{R}^3)$ has Fourier transform \begin{align*} \hat{u}(\xi) &= \frac{\hat{f}(\xi)}{1+|\xi|^2} = \frac{-|\xi|^2\hat{g}(\xi)}{1+|\xi|^2}. \end{align*} Using the partial fraction decomposition \begin{align*} \frac{|\xi|^2}{1+|\xi|^2} &= 1 - \frac{1}{1+|\xi|^2}, \end{align*} we get $\hat{u} = -\hat{g} + (1+|\xi|^2)^{-1}\hat{g}$, giving \begin{align*} u &= -g + (1-\Delta)^{-1}g. \end{align*} The first term is in $H^1$ and the second in $H^3$ (by the [Helmholtz isomorphism](/theorems/227) applied to $g \in H^1$), confirming $u \in H^1$ with $\|u\|_{H^1} \le 2\|g\|_{H^1}$. [/example] More generally, the Bessel potential operator $(1-\Delta)^{s/2}$, defined on the Fourier side as multiplication by $(1+|\xi|^2)^{s/2}$, gives an isometric isomorphism \begin{align*} (1-\Delta)^{s/2}: H^t(\mathbb{R}^n) &\xrightarrow{\;\sim\;} H^{t-s}(\mathbb{R}^n) \end{align*} for every $t, s \in \mathbb{R}$. This gives the $H^s$ scale the structure of a "ruler" for regularity: applying $(1-\Delta)^{s/2}$ shifts the index by exactly $s$. ## The Algebra Property and Nonlinear PDEs For nonlinear PDEs, the fundamental structural question is whether $H^s$ is closed under pointwise multiplication. If $u, v \in H^s$, is the product $uv$ again in $H^s$? The answer depends on whether $s$ exceeds the embedding threshold. [quotetheorem:468] The threshold $s > n/2$ is the same as for the [Sobolev embedding](/theorems/226), and this is no coincidence: the proof for integer $s = k$ uses the Leibniz rule $\partial^\alpha(fg) = \sum_{\beta \le \alpha}\binom{\alpha}{\beta}(\partial^\beta f)(\partial^{\alpha-\beta}g)$ and bounds each term by placing the lower-order factor in $L^\infty$ (via the embedding) and the higher-order factor in $L^2$. The threshold is sharp: for $s \le n/2$, one can construct $f, g \in H^s$ with $fg \notin H^s$ (using functions with logarithmic singularities). [example: The Algebra Property In Action] For $n = 3$ and $s = 2 > 3/2 = n/2$, the algebra property gives $\|fg\|_{H^2} \le C\|f\|_{H^2}\|g\|_{H^2}$ for all $f, g \in H^2(\mathbb{R}^3)$. As a concrete check, take $f = g = e^{-|x|^2}$. Then $fg = e^{-2|x|^2}$ is another Gaussian, and \begin{align*} \|fg\|_{H^2}^2 &= \int_{\mathbb{R}^3} (1+|\xi|^2)^2 \, |\widehat{e^{-2|\cdot|^2}}(\xi)|^2 \, d\mathcal{L}^3(\xi). \end{align*} Since $\widehat{e^{-a|x|^2}}(\xi) = (\pi/a)^{3/2}e^{-\pi^2|\xi|^2/a}$, this is a finite Gaussian integral, confirming $fg \in H^2$. [/example] [example: Local Well-Posedness For A Nonlinear Wave Equation] The algebra property is the key input for local well-posedness of nonlinear PDEs. Consider the semilinear [wave equation](/page/Wave%20Equation) on $\mathbb{R}^{1+n}$: \begin{align*} \partial_{tt}u - \Delta u &= u^3, \\ u(0,\cdot) &= u_0, \quad \partial_t u(0,\cdot) = u_1, \end{align*} with $(u_0, u_1) \in H^s(\mathbb{R}^n) \times H^{s-1}(\mathbb{R}^n)$ for $s > n/2$. The solution is a fixed point of $\Phi: u \mapsto w$, where $w$ solves the linear wave equation with forcing $u^3$. The energy estimate gives \begin{align*} \|w(t)\|_{H^s} + \|\partial_t w(t)\|_{H^{s-1}} &\lesssim \|u_0\|_{H^s} + \|u_1\|_{H^{s-1}} + \int_0^t\|u(\tau)^3\|_{H^{s-1}}\,d\tau. \end{align*} The [algebra property](/theorems/468) controls the nonlinearity: \begin{align*} \|u^3\|_{H^s} &\le C_{n,s}^2\|u\|_{H^s}^3. \end{align*} For $T$ small enough depending on $\|u_0\|_{H^s} + \|u_1\|_{H^{s-1}}$, the map $\Phi$ is a contraction on a ball in $C([0,T]; H^s) \times C([0,T]; H^{s-1})$, yielding a unique local solution. The threshold $s > n/2$ is essential — for $s \le n/2$ the algebra property fails and the fixed-point argument breaks down. [/example] ## Relation to Homogeneous Spaces The [homogeneous Sobolev space](/page/Homogeneous%20Sobolev%20Space) $\dot{H}^s(\mathbb{R}^n)$ uses the weight $|\xi|^{2s}$ instead of $(1+|\xi|^2)^s$. Since $(1+|\xi|^2)^s$ dominates $|\xi|^{2s}$ at all frequencies (the "$+1$" provides low-frequency control), the inhomogeneous norm dominates the homogeneous one. The question is whether this comparison gives a continuous embedding. [quotetheorem:467] The embedding is strict: $\dot{H}^s(\mathbb{R}^n)$ contains elements with no $L^2$ control. For instance, constant functions have $|\xi|^s\hat{c} = c \cdot |\xi|^s\delta_0$, which vanishes for $s > 0$ in the sense that $\|c\|_{\dot{H}^s} = 0$ — so constants belong to $\dot{H}^s/\mathcal{P}$. But \begin{align*} \|c\|_{H^0}^2 &= \int_{\mathbb{R}^n} |c|^2\,d\mathcal{L}^n = \infty \quad \text{for } c \ne 0, \end{align*} so constants are not in $H^0 \cong L^2$. The inhomogeneous space is the right setting whenever both the function and its derivatives must be controlled simultaneously — for elliptic equations on $\mathbb{R}^n$ and Cauchy problems requiring finite-energy initial data. ## References 1. H. Brezis, *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2010). 2. L. C. Evans, *Partial Differential Equations* (1998). 3. M. E. Taylor, *Partial Differential Equations III: Nonlinear Equations* (1997). 4. T. Tao, *Nonlinear Dispersive Equations: Local and Global Analysis* (2006). 5. E. M. Stein, *Singular Integrals and Differentiability Properties of Functions* (1970).