Hadamard's lemma

In mathematics, Hadamard's lemma, named after Jacques Hadamard, is essentially a first-order form of Taylor's theorem, in which we can express a smooth, real-valued function exactly in a convenient manner.

Statement

Hadamard's lemma — Let $f$ be a smooth, real-valued function defined on an open, star-convex neighborhood $U$ of a point $a$ in $n$-dimensional Euclidean space. Then $f(x)$ can be expressed, for all $x\in U,$ in the form:

$$f(x)=f(a)+\sum _{i=1}^{n}\left(x_{i}-a_{i}\right)g_{i}(x),$$

where each $g_{i}$ is a smooth function on $U,$ $a=\left(a_{1},\ldots ,a_{n}\right),$ and $x=\left(x_{1},\ldots ,x_{n}\right).$

Proof

Proof

Let $x\in U.$ Define $h:[0,1]\to \mathbb {R}$ by

$$h(t)=f(a+t(x-a))\qquad {\text{ for all }}t\in [0,1].$$

Then

$$h'(t)=\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\left(x_{i}-a_{i}\right),$$

which implies

$${\begin{aligned}h(1)-h(0)&=\int _{0}^{1}h'(t)\,dt\\&=\int _{0}^{1}\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\left(x_{i}-a_{i}\right)\,dt\\&=\sum _{i=1}^{n}\left(x_{i}-a_{i}\right)\int _{0}^{1}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\,dt.\end{aligned}}$$

But additionally, $h(1)-h(0)=f(x)-f(a),$ so by letting

$$g_{i}(x)=\int _{0}^{1}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\,dt,$$

the theorem has been proven. $\blacksquare$

Consequences and applications

Corollary — If $f:\mathbb {R} \to \mathbb {R}$ is smooth and $f(0)=0$ then $f(x)/x$ is a smooth function on $\mathbb {R} .$ Explicitly, this conclusion means that the function $\mathbb {R} \to \mathbb {R}$ that sends $x\in \mathbb {R}$ to

$${\begin{cases}f(x)/x&{\text{ if }}x\neq 0\\\lim _{t\to 0}f(t)/t&{\text{ if }}x=0\\\end{cases}}$$

is a well-defined smooth function on $\mathbb {R} .$

Proof

By Hadamard's lemma, there exists some $g\in C^{\infty }(\mathbb {R} )$ such that $f(x)=f(0)+xg(x)$ so that $f(0)=0$ implies $f(x)/x=g(x).$ $\blacksquare$

Corollary — If $y,z\in \mathbb {R} ^{n}$ are distinct points and $f:\mathbb {R} ^{n}\to \mathbb {R}$ is a smooth function that satisfies $f(z)=0=f(y)$ then there exist smooth functions $g_{i},h_{i}\in C^{\infty }\left(\mathbb {R} ^{n}\right)$ ($i=1,\ldots ,3n-2$) satisfying $g_{i}(z)=0=h_{i}(y)$ for every $i$ such that

$$f=\sum _{i}^{}g_{i}h_{i}.$$

Proof

By applying an invertible affine linear change in coordinates, it may be assumed without loss of generality that $z=(0,\ldots ,0)$ and $y=(0,\ldots ,0,1).$ By Hadamard's lemma, there exist $g_{1},\ldots ,g_{n}\in C^{\infty }\left(\mathbb {R} ^{n}\right)$ such that $f(x)=\sum _{i=1}^{n}x_{i}g_{i}(x).$ For every $i=1,\ldots ,n,$ let $\alpha _{i}:=g_{i}(y)$ where $0=f(y)=\sum _{i=1}^{n}y_{i}g_{i}(y)=g_{n}(y)$ implies $\alpha _{n}=0.$ Then for any $x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n},$

$${\begin{alignedat}{8}f(x)&=\sum _{i=1}^{n}x_{i}g_{i}(x)&&\\&=\sum _{i=1}^{n}\left[x_{i}\left(g_{i}(x)-\alpha _{i}\right)\right]+\sum _{i=1}^{n-1}\left[x_{i}\alpha _{i}\right]&&\quad {\text{ using }}g_{i}(x)=\left(g_{i}(x)-\alpha _{i}\right)+\alpha _{i}{\text{ and }}\alpha _{n}=0\\&=\left[\sum _{i=1}^{n}x_{i}\left(g_{i}(x)-\alpha _{i}\right)\right]+\left[\sum _{i=1}^{n-1}x_{i}x_{n}\alpha _{i}\right]+\left[\sum _{i=1}^{n-1}x_{i}\left(1-x_{n}\right)\alpha _{i}\right]&&\quad {\text{ using }}x_{i}=x_{n}x_{i}+x_{i}\left(1-x_{n}\right).\\\end{alignedat}}$$

Each of the $3n-2$ terms above has the desired properties. $\blacksquare$

See also

• Bump function – Smooth and compactly supported function
• Continuously differentiable – Mathematical function whose derivative existsPages displaying short descriptions of redirect targets
• Smoothness – Number of derivatives of a function (mathematics)
• Taylor's theorem – Approximation of a function by a truncated power series

References

• Nestruev, Jet (2002). Smooth manifolds and observables. Berlin: Springer. ISBN 0-387-95543-7.
• Nestruev, Jet (10 September 2020). Smooth Manifolds and Observables. Graduate Texts in Mathematics. Vol. 220. Cham, Switzerland: Springer Nature. ISBN 978-3-030-45649-8. OCLC 1195920718.