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.

tatement

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 for "x" in "U", we have

:$f(x)=f(a)+sum\_\{i=1\}^n\; (x\_i-a\_i)\; g\_i(x),$

where each "g_i" is a smooth function on "U", "a" = ("a"₁,...,"a"_"n"), and "x" = ("x"₁,...,"x"_"n").

Proof

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

:$h(t)=f(a+t(x-a)).,$

Then since

:$h\text{'}(t)=sum\_\{i=1\}^n\; frac\{partial\; f\}\{partial\; x\_i\}(x+t(x-a))\; (x\_i-a\_i),$

we have

:$h(1)-h(0)=int\_0^1\; h\text{'}(t),dt=int\_0^1\; sum\_\{i=1\}^n\; frac\{partial\; f\}\{partial\; x\_i\}(x+t(x-a))\; (x\_i-a\_i),\; dt=sum\_\{i=1\}^n\; (x\_i-a\_i)int\_0^1\; frac\{partial\; f\}\{partial\; x\_i\}(a+t(x-a)),\; dt.$

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

:$g\_i(x)=int\_0^1\; frac\{partial\; f\}\{partial\; x\_i\}(a+t(x-a)),\; dt,$

we have proven the theorem.

References