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Let $D_1$ and $D_2$ each be a connection on a vector bundle. Their difference is tensorial, so $D_1=D_2+a$ with $a\in {\Omega }^1(\mathrm{End}E)$. Extend $D_2$ to the endomorphism bundle connection. Expanding $(D_2+a{)}^2$ with the supplied graded Leibniz rule gives the curvature difference formula

For the $R$-$R$-bimodule $M$, the Hochschild chain complex iswith boundaryThenThe Hochschild cochain complex is $C^n(R,M)={\mathrm{Hom}}_k(R^{{\otimes }_{k}n},M)$ withand $H{H}^n(R,M)=H^n(C^{\bullet }(R,M),\delta )$.

A derivation into a bimodule is a $k$-linear map $d:R\to M$ satisfyingIt is an inner derivation when $d=d_m$ for some $m\in M$, where $d_m(r)=rm-mr$. The degree-one cocycle equation is precisely the Leibniz rule, and the degree-one coboundaries are the inner derivations. Therefore the First Hochschild cohomology as outer derivations is

Let $\mu :R{\otimes }_{k}R\to R$ be multiplication and $\Omega =\ker \mu$, equipped with the outer bimodule structure. The universal bimodule derivationsatisfies $D(rs)=rD(s)+D(r)s$. Composition with $D$ definesFor a derivation $d$, its inverse image under this map isIf ${\sum }_i{r}_i{s}_i=0$, the Leibniz rule shows that this formula is both left and right $R$-linear. Moreover every element of $\Omega$ is ${\sum }_{i}D(r_i)s_i$, proving existence and uniqueness.

For $d_m(r)=rm-mr$, the corresponding map isThese are exactly the maps $\Omega \to M$ which extend to bimodule maps $R{\otimes }_{k}R\to M$.

The Leibniz rule rewrites the left side as $(xu{)}^{\prime }$. By a distribution with zero derivative is constant,The Dirac delta multiplication identity $x{\delta }_0^{\prime }=-{\delta }_0$ supplies a particular solution, and the preceding division result supplies all solutions of $xu=c$. HenceMultiplying by $x$ and then differentiating verifies the equation. The coordinate-multiplication kernel and the constant-derivative kernel show that no additional solutions are missing.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted Leibniz ruleThe Zariski tangent space isSince $p(f)\ne 0$, $p$ extends uniquely to the localization $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f{)}^n$. Restriction gives a linear mapConversely, every $D\in T_{p}A$ has the unique extensionThe quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.

Let $Q_k=P^{(k)}$. Differentiating Legendre's differential equation $k$ times and using the Leibniz rule givesThe induction step follows by differentiating this equation once: the derivative of the coefficient $-2(k+1)x$ contributes the additional term $-2(k+1)Q_k^{\prime }$. ThusFor fixed $\lambda$, this is negative for every sufficiently large $k$.

By the Leibniz rule,After multiplication by $(-1{)}^n{x}^{-1/2}{e}^x$, the term $j=0$ is $x^n$, and every other term has lower degree. Thus $p_n$ is a monic polynomial of degree $n$. Direct calculation givesThese are the monic normalization of the generalized Laguerre polynomials with parameter $1/2$.

The multivariable Leibniz rule writes each derivative of $a_1{a}_2$ as a finite sum of productsEach term is bounded by a constant timesHence

Applying the distributional Leibniz rule twice givesEliminating the middle term provesChoose $\rho \in \mathcal{D}(\mathbb{R})$ with $\rho (0)=1$. Parts i and ii and the identity just proved, with $\phi =\rho$ and $u=xH$, yieldThus explicit continuous functions of compact support are

The distributional derivative of the Heaviside step function is $H^{\prime }={\delta }_0$. The Leibniz rule and $x{\delta }_0=0$ therefore give

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

The Leibniz rule writes every derivative of $a_1{a}_2$ as a finite sum of products of derivatives of the two factors. Multiplying the corresponding symbol estimates adds their orders, so

A Riemannian metric $g$ on a complex manifold is a Kähler metric when its complex-linear extension is a Hermitian form on each tangent space and its fundamental two-formis closed. Equivalently, $\omega$ is a positive real closed $(1,1)$-form, making $X$ a Kähler manifold.

The Lefschetz operator of a Kähler manifold and its adjoint areBecause $d\omega =0$ and $\omega$ has type $(1,1)$, both $\partial \omega$ and $\bar{\partial }\omega$ vanish. The graded Leibniz rule therefore gives $[L,\partial ]=[L,\bar{\partial }]=0$.

Writing formal adjoints with stars, define the three Laplacians byThe supplied Kähler identities identity $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives, by complex conjugation and taking adjoints,Expanding these commutators and using $\partial \bar{\partial }+\bar{\partial }\partial =0$ shows that the mixed terms in $\Delta$ vanish and that ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Since $d=\partial +\bar{\partial }$, it follows that

The adjoint of $[L,\partial ]=0$ gives $[\Lambda ,{\partial }^{\ast }]=0$. Hencewhere the last equality is the adjoint of $\partial \bar{\partial }+\bar{\partial }\partial =0$. Thus $\Lambda$, and therefore every power ${\Lambda }^k$, commutes with ${\Delta }_{\partial }$. It follows that ${\Lambda }^k$ sends every $\partial$-harmonic $(p,q)$-form to a $\partial$-harmonic $(p-k,q-k)$-form whenever $0\le k\le \min (p,q)$.

True. Repeated use of the Leibniz rule and the polynomial Taylor formula gives

The Leibniz rule writes every derivative of $a_1{a}_2$ as a finite sum of products of derivatives of the factors. Multiplying their symbol bounds adds the orders, so

The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^n;N)=S_{1,0}^N$ consists of $c\in C^{\infty }(X\times {\mathbb{R}}^n)$ such that, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,The defining estimate givesThe Leibniz rule givesand the triangle inequality givesThese are the basic rules of symbol calculus.

For a Riemannian metric on an ambient manifold and an embedded submanifold, let $\pi :TN{|}_M\to TM$ be orthogonal projection. Project the restriction of a connection to an embedded submanifold to obtain $D_{X}Y=\pi {\nabla }_{X}Y$. Since $\pi Y=Y$, the Leibniz rule and linearity over smooth functions hold, making $D$ an affine connection. If the ambient connection is the Levi-Civita connection, pairing with tangent vectors shows that $D$ is a metric connection; projecting the zero-torsion identity shows it is a torsion-free connection. By uniqueness, $D$ is the induced Levi-Civita connection.

For a holomorphic subbundle $S\subset E$ and quotient $Q=E/S$, the second fundamental form is $A(s)=(I-{\pi }_S)D_{E}s$. The Leibniz rule gives $A(fs)=fA(s)$, and holomorphicity of the subbundle makes its $(0,1)$ part vanish. Thus $A\in {\mathcal{A}}^{1,0}(\mathrm{Hom}(S,Q))$: the form factor is a cotangent bundle, not a tangent bundle.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

For an integer $k\ge 0$ and sufficiently regular bounded functions, $\Vert fg{\Vert }_{H^k}\le C_k(\Vert f{\Vert }_{\infty }\Vert g{\Vert }_{H^k}+\Vert g{\Vert }_{\infty }\Vert f{\Vert }_{H^k})$. The Leibniz rule and derivative interpolation give the estimate. In a differentiated semilinear wave equation, this separates a controlled low-order coefficient from an arbitrarily high Sobolev norm, allowing propagation of smoothness after a fixed base norm has been controlled.