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

Let ƒ be a smooth, real-valued function defined on an open, star-convex neighborhood U of a point a in n-dimensional Euclidean space. Then ƒ(x) can be expressed, for all 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 g_i is a smooth function on U, a = (a₁, …, a_n), and x = (x₁, …, x_n).

Proof

Let x be in U. Let h be the map from [0,1] to the real numbers defined by

${\displaystyle h(t)=f(a+t(x-a)).}$

Then since

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

we have

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

But, additionally, h(1) − h(0) = f(x) − f(a), so if we let

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

we have proven the theorem.

References

• Nestruev, Jet (2002). Smooth manifolds and observables. Berlin: Springer. ISBN 0-387-95543-7.