Hadamard's lemma

Statement

Hadamard's lemma^[1] — Let ${\displaystyle f}$ be a smooth, real-valued function defined on an open, star-convex neighborhood ${\displaystyle U}$ of a point ${\displaystyle a}$ in ${\displaystyle n}$-dimensional Euclidean space. Then ${\displaystyle f(x)}$ can be expressed, for all ${\displaystyle x\in U,}$ in the form: $${\displaystyle f(x)=f(a)+\sum _{i=1}^{n}\left(x_{i}-a_{i}\right)g_{i}(x),}$$ where each ${\displaystyle g_{i}}$ is a smooth function on ${\displaystyle U,}$ ${\displaystyle a=\left(a_{1},\ldots ,a_{n}\right),}$ and ${\displaystyle x=\left(x_{1},\ldots ,x_{n}\right).}$

Proof

Let ${\displaystyle x\in U.}$ Define ${\displaystyle h:[0,1]\to \mathbb {R} }$ by $${\displaystyle h(t)=f(a+t(x-a))\qquad {\text{ for all }}t\in [0,1].}$$

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

But additionally, ${\displaystyle h(1)-h(0)=f(x)-f(a),}$ so by letting $${\displaystyle g_{i}(x)=\int _{0}^{1}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\,dt,}$$ the theorem has been proven. ${\displaystyle \blacksquare }$

Consequences and applications

Corollary^[1] — If ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is smooth and ${\displaystyle f(0)=0}$ then ${\displaystyle f(x)/x}$ is a smooth function on ${\displaystyle \mathbb {R} .}$ Explicitly, this conclusion means that the function ${\displaystyle \mathbb {R} \to \mathbb {R} }$ that sends ${\displaystyle x\in \mathbb {R} }$ to $${\displaystyle {\begin{cases}f(x)/x&{\text{ if }}x\neq 0\\\lim _{t\to 0}f(t)/t&{\text{ if }}x=0\\\end{cases}}}$$ is a well-defined smooth function on ${\displaystyle \mathbb {R} .}$

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

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

By applying an invertible affine linear change in coordinates, it may be assumed without loss of generality that ${\displaystyle z=(0,\ldots ,0)}$ and ${\displaystyle y=(0,\ldots ,0,1).}$ By Hadamard's lemma, there exist ${\displaystyle g_{1},\ldots ,g_{n}\in C^{\infty }\left(\mathbb {R} ^{n}\right)}$ such that ${\displaystyle f(x)=\sum _{i=1}^{n}x_{i}g_{i}(x).}$ For every ${\displaystyle i=1,\ldots ,n,}$ let ${\displaystyle \alpha _{i}:=g_{i}(y)}$ where ${\displaystyle 0=f(y)=\sum _{i=1}^{n}y_{i}g_{i}(y)=g_{n}(y)}$ implies ${\displaystyle \alpha _{n}=0.}$ Then for any ${\displaystyle x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n},}$ $${\displaystyle {\begin{alignedat}{8}f(x)&=\sum _{i=1}^{n}x_{i}g_{i}(x)&&\\&=\sum _{i=1}^{n}\left[x_{i}\left(g_{i}(x)-\alpha _{i}\right)\right]+\sum _{i=1}^{n-1}\left[x_{i}\alpha _{i}\right]&&\quad {\text{ using }}g_{i}(x)=\left(g_{i}(x)-\alpha _{i}\right)+\alpha _{i}{\text{ and }}\alpha _{n}=0\\&=\left[\sum _{i=1}^{n}x_{i}\left(g_{i}(x)-\alpha _{i}\right)\right]+\left[\sum _{i=1}^{n-1}x_{i}x_{n}\alpha _{i}\right]+\left[\sum _{i=1}^{n-1}x_{i}\left(1-x_{n}\right)\alpha _{i}\right]&&\quad {\text{ using }}x_{i}=x_{n}x_{i}+x_{i}\left(1-x_{n}\right).\\\end{alignedat}}}$$ Each of the ${\displaystyle 3n-2}$ terms above has the desired properties. ${\displaystyle \blacksquare }$

See also

Citations

References