مأخوذة هادامار

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: $$f(x)=f(a)+\sum _{i=1}^n(x_i-a_i)g_i(x),$$ where each $g_i$ is a smooth function on $U,$ $a=(a_1,\dots ,a_n),$ and $x=(x_1,\dots ,x_n).$

 Proof

Proof

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

Then $$h^{\prime }(t)=\sum _{i=1}^n\frac{\partial f}{\partial x_i}(a+t(x-a))(x_i-a_i),$$ which implies $$\begin{array}{ll}h(1)-h(0)& =\underset{0}{\overset{1}{\int }}h\text{'}(t)dt\\ & =\underset{0}{\overset{1}{\int }}\sum _{i=1}^n\frac{\partial f}{\partial x_i}(a+t(x-a))(x_i-a_i)dt\\ & =\sum _{i=1}^n(x_i-a_i)\underset{0}{\overset{1}{\int }}\frac{\partial f}{\partial x_i}(a+t(x-a))dt.\end{array}$$

But additionally, $h(1)-h(0)=f(x)-f(a),$ so by letting $$g_i(x)=\underset{0}{\overset{1}{\int }}\frac{\partial f}{\partial x_i}(a+t(x-a))dt,$$ the theorem has been proven. $◼$

 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 $$\{\begin{array}{cc}f(x)/x& \text{ if }x\ne 0\\ \underset{t\to 0}{\lim }f(t)/t& \text{ if }x=0\\ \end{array}$$ is a well-defined smooth function on $\mathbb{ℝ}.$

Proof

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

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 $g_i,h_i\in C^{\mathrm{\infty }}\left({\mathbb{ℝ}}^n\right)$ ($i=1,\dots ,3n-2$) 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 $z=(0,\dots ,0)$ and $y=(0,\dots ,0,1).$ By Hadamard's lemma, there exist $g_1,\dots ,g_n\in C^{\mathrm{\infty }}\left({\mathbb{ℝ}}^n\right)$ such that $f(x)=\sum _{i=1}^n{x}_i{g}_i(x).$ For every $i=1,\dots ,n,$ 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 $x=(x_1,\dots ,x_n)\in {\mathbb{ℝ}}^n,$ $$\begin{array}{rrrr}f(x)& =\sum _{i=1}^n{x}_i{g}_i(x)& & \\ & =\sum _{i=1}^n\left[x_i(g_i(x)-{\alpha }_i)\right]+\sum _{i=1}^{n-1}\left[x_i{\alpha }_i\right]& & \text{ using }{g}_i(x)=(g_i(x)-{\alpha }_i)+{\alpha }_i\text{ and }{\alpha }_n=0\\ & =\left[\sum _{i=1}^n{x}_i(g_i(x)-{\alpha }_i)\right]+\left[\sum _{i=1}^{n-1}{x}_i{x}_n{\alpha }_i\right]+\left[\sum _{i=1}^{n-1}{x}_i(1-x_n){\alpha }_i\right]& & \text{ using }{x}_i=x_n{x}_i+x_i(1-x_n).\\ \end{array}$$ Each of the $3n-2$ terms above has the desired properties. $◼$

 See also

• Bump function
• Continuously differentiable
• Smoothness
• Taylor's theorem

 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.