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The H1 space is the first-order Sobolev space whose elements and first weak derivatives belong to L2 space. It is a Hilbert space with inner product given by the sum of the L2 inner products of the functions and of their first weak derivatives.

On $I=(-a,a)$, a first-order Sobolev space function has a continuous representative and obeysBound the average by the Cauchy-Schwarz inequality and average the identity $u(x)-u(y)={\int }_y^x{u}^{\prime }$. This also gives $[u{]}_{C^{0,1/2}}\le \Vert u^{\prime }{\Vert }_2$. A unit constant with the standard Sobolev norm cannot be asserted on arbitrary interval lengths: $u=1$ already disproves it when $2a<1$.

Let $Mf$ be the uncentered maximal function of $|f|$, defined by its averages over balls. First establish the strong $L^p$ estimate needed below. The uncentered maximal weak-type inequality gives the weak $L^1$ bound with $C_d=3^d$, and plainly $Mf\le \Vert f{\Vert }_{\infty }$ for bounded inputs. For $f\in L^p$, split it into its parts on $\{|f|>a/2\}$ and $\{|f|\le a/2\}$. The first part is in $L^1$, since $|f|1_{\{|f|>a/2\}}\le (a/2{)}^{1-p}|f{|}^p$, and the second part contributes at most $a/2$ to the maximal function. Sublinearity therefore givesIntegrating this bound with layer cake representation and Tonelli theorem yieldsThis proves the strong Lp bound from weak L1 and L-infinity bounds directly. In particular $Mf$ is finite almost everywhere.

Now split the positive Riesz potential of $|f|$ at a radius $r$. In the near region, decompose into annuli $2^{-j-1}r<|x-y|\le 2^{-j}r$. On the $j$th annulus,The geometric series converges because $\alpha >0$. In the far region, Hölder's inequality, with $p^{\prime }=p/(p-1)$, givesThe radial integral is finite exactly because $p<d/\alpha$. Combining these bounds, including the fixed Riesz normalization constant, givesFor $0<Mf(x)<\infty$, choose $r=(\Vert f{\Vert }_p/Mf(x){)}^{p/d}$. If $Mf(x)=0$, every ball average vanishes and $f=0$ almost everywhere, so the assertion is trivial. The zero-norm case is likewise trivial. The resulting Hedberg inequality isThis proves absolute convergence of the defining integral for almost every $x$, also for signed or complex $f$. Put $\theta =\alpha p/d\in (0,1)$. The proposed exponent satisfies $q=p/(1-\theta )$, so $(1-\theta )q=p$. Raising the pointwise estimate to the $q$th power and using the proved maximal bound givesThus the Hardy–Littlewood–Sobolev inequality follows from the estimates just derived, rather than being quoted in place of its proof.

For uniqueness of the target exponent, suppose such a uniform estimate holds with index $r$ instead. Choose a nonzero nonnegative bump function $f$ and let $f_t(x)=f(tx)$ for $t>0$. Homogeneity of the Riesz potential givesThe assumed estimate at $t=1$ makes the nonzero output's $L^r$ norm finite. For all $t>0$ it must therefore satisfywith $d/r=0$ when $r=\infty$. Letting $t$ tend to zero or infinity forces the power of $t$ to vanish:Hence no other target index is possible.

For the Sobolev embedding, take $d>1$ and $1<p<d$, so the preceding result is available with $\alpha =1$. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the vector field $(x-y)/|x-y{|}^d$ has zero divergence in $y$ away from $x$. Integrating by parts outside the ball $|x-y|\le \epsilon$, its inner boundary contributes $|{\mathbb{S}}^{d-1}|$ times the spherical average of $u$ about $x$. The outer boundary contributes zero because $u$ has compact support. Letting $\epsilon \downarrow 0$ provesThe absolute value is at most $C{I}_1(|\nabla u|)(x)$. ThusFor a general member of the first-order Sobolev space $W^{1,p}({\mathbb{R}}^d)=L_1^p({\mathbb{R}}^d)$, approximate in the Sobolev norm by compactly supported smooth functions. This density of smooth functions in a Sobolev space follows from cutoffs tending to one and subsequent mollification: the cutoff-gradient term tends to zero in $L^p$, and a mollifier approximates both the function and each weak derivative. The displayed estimate makes the approximants Cauchy in $L^{q_1}$. Their $L^p$ limit is the original function, so subsequences converging almost everywhere identify the $L^{q_1}$ limit with it. This proves continuous embedding. The range $1<p<d$ is essential for this consequence; the formula is not a general critical $p=d$ embedding into $L^{\infty }$.

For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the fundamental theorem of calculus along the $j$th coordinate line givesTake $f=|u|$ and $s_j=|{\partial }_{j}u|$ in the introductory result. It follows thatThe last step is the arithmetic-geometric mean inequality. This bounds the target norm by the first-order Sobolev space norm. Approximate any $u\in W^{1,1}({\mathbb{R}}^d)=L_1^1({\mathbb{R}}^d)$ by compactly supported smooth functions, using cutoffs and mollification as in Question 4. The estimate applied to differences gives convergence in $L^{d/(d-1)}$, and subsequences of the simultaneous $L^1$ and target-norm convergence identify the two limits almost everywhere. Thus $W^{1,1}$ embeds continuously into $L^{d/(d-1)}$. The coordinate-derivative product bound also passes to the limit. For $d=1$, the line argument instead gives $\Vert u{\Vert }_{\infty }\le \Vert u^{\prime }{\Vert }_1$ and the embedding into $L^{\infty }$.

Use the angular-frequency Fourier transform conventionA precise sufficient interpretation of the regularity assumption is $f\ne 0$, $f\in H^1(\mathbb{R})$ in the first-order Sobolev space and $xf\in L^2(\mathbb{R})$; in particular, nonzero Schwartz functions are admissible. We use the Plancherel theorem, in the normalization $\Vert \hat{f}{\Vert }_2^2=2\pi \Vert f{\Vert }_2^2$, together with its derivative identity $\widehat{f^{\prime }}(\lambda )=i\lambda \hat{f}(\lambda )$. Both identities hold in $L^2$ for $f\in H^1$, by extending their identities for Schwartz functions. Thus the frequency factor isFor a Schwartz function, integration by parts gives $\Vert f{\Vert }_2^2=-2\mathrm{Re}\int x{f}^{\prime }(x)\overline{f(x)}dx$. This remains true under the stated assumptions: insert a smooth cutoff ${\chi }_R(x)=\chi (x/R)$ into the derivative of $x|f{|}^2$, where $\chi$ is one near zero and compactly supported. Integration givesThe cutoff-derivative term tends to zero, since $x{\chi }_R^{\prime }$ is bounded independently of $R$ and supported in a tail. The other terms converge by dominated convergence and the integrability of $|f^{\prime }||xf|$, furnished by the Cauchy-Schwarz inequality. HenceAfter squaring and using the Plancherel theorem, we obtain the uncentred Fourier uncertainty principle:If equality holds, both displayed inequalities must be equalities. The equality condition in the Cauchy-Schwarz inequality gives $f^{\prime }=cxf$ almost everywhere for a constant $c\in \mathbb{C}$; the real-part inequality and the identity for $\Vert f{\Vert }_2^2$ then force $c=-a$ with $a>0$ real. The weak derivative equation has the solution $f(x)=C{e}^{-a{x}^2/2}$: locally multiply by $e^{a{x}^2/2}$ to obtain a function whose weak derivative is zero, hence a constant. Conversely, for this Gaussian function the normalized second moments are $1/(2a)$ in position and $a/2$ in angular frequency, whose product is $1/4$. ThereforeThese are raw second moments, rather than variances about their respective means. Consequently translations and nonzero frequency modulations of the Gaussian function are not equality cases here; the equality case of the Heisenberg uncertainty relation for centred variances has those additional freedoms.

For $I=(-a,a)$, the first-order Sobolev space isThe weak derivative $v=u^{\prime }$ is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm $\Vert u{\Vert }_{H^1(I)}^2=\Vert u{\Vert }_{L^2(I)}^2+\Vert u^{\prime }{\Vert }_{L^2(I)}^2$. Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.

For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

For $f\in H^1(\mathbb{R})$, its difference quotient equals $D_{h}f={\int }_0^1{f}^{\prime }(\cdot +\theta h)d\theta$ in L2 space. Thus $\Vert D_{h}f{\Vert }_2\le \Vert f^{\prime }{\Vert }_2$, and translation of a function gives $D_{h}f\to f^{\prime }$ in L2 space. Conversely, bounded difference quotients have a weakly convergent subsequence as $h\to 0$. Against a test function, $\int D_{h}fv\to -\int f{v}^{\prime }$, so that weak limit is the weak derivative of $f$. It lies in L2 space, which is precisely the first-order Sobolev space condition.