Hadamard's lemma

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

Proof

Proof

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

Then h ′ ( t ) = ∑ i = 1 n ∂ f ∂ x i ( a + t ( x − a ) ) ( x i − a i ) , {\displaystyle h'(t)=\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\left(x_{i}-a_{i}\right),} {\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 h ( 1 ) − h ( 0 ) = ∫ 0 1 h ′ ( t ) d t = ∫ 0 1 ∑ i = 1 n ∂ f ∂ x i ( a + t ( x − a ) ) ( x i − a i ) d t = ∑ i = 1 n ( x i − a i ) ∫ 0 1 ∂ f ∂ x i ( a + t ( x − a ) ) d t . {\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}}} {\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, h ( 1 ) − h ( 0 ) = f ( x ) − f ( a ) , {\displaystyle h(1)-h(0)=f(x)-f(a),} {\displaystyle h(1)-h(0)=f(x)-f(a),} so by letting g i ( x ) = ∫ 0 1 ∂ f ∂ x i ( a + t ( x − a ) ) d t , {\displaystyle g_{i}(x)=\int _{0}^{1}{\frac {\partial f}{\partial x_{i}}}(a+t(x-a))\,dt,} {\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 } {\displaystyle \blacksquare }

Consequences and applications

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

Proof

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

Corollary[1]—If y , z ∈ R n {\displaystyle y,z\in \mathbb {R} ^{n}} {\displaystyle y,z\in \mathbb {R} ^{n}} are distinct points and f : R n → R {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} } {\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} } is a smooth function that satisfies f ( z ) = 0 = f ( y ) {\displaystyle f(z)=0=f(y)} {\displaystyle f(z)=0=f(y)} then there exist smooth functions g i , h i ∈ C ∞ ( R n ) {\displaystyle g_{i},h_{i}\in C^{\infty }\left(\mathbb {R} ^{n}\right)} {\displaystyle g_{i},h_{i}\in C^{\infty }\left(\mathbb {R} ^{n}\right)} ( i = 1 , … , 3 n − 2 {\displaystyle i=1,\ldots ,3n-2} {\displaystyle i=1,\ldots ,3n-2}) satisfying g i ( z ) = 0 = h i ( y ) {\displaystyle g_{i}(z)=0=h_{i}(y)} {\displaystyle g_{i}(z)=0=h_{i}(y)} for every i {\displaystyle i} {\displaystyle i} such that f = ∑ i g i h i . {\displaystyle f=\sum _{i}^{}g_{i}h_{i}.} {\displaystyle 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 , … , 0 ) {\displaystyle z=(0,\ldots ,0)} {\displaystyle z=(0,\ldots ,0)} and y = ( 0 , … , 0 , 1 ) . {\displaystyle y=(0,\ldots ,0,1).} {\displaystyle y=(0,\ldots ,0,1).} By Hadamard's lemma, there exist g 1 , … , g n ∈ C ∞ ( R n ) {\displaystyle g_{1},\ldots ,g_{n}\in C^{\infty }\left(\mathbb {R} ^{n}\right)} {\displaystyle g_{1},\ldots ,g_{n}\in C^{\infty }\left(\mathbb {R} ^{n}\right)} such that f ( x ) = ∑ i = 1 n x i g i ( x ) . {\displaystyle f(x)=\sum _{i=1}^{n}x_{i}g_{i}(x).} {\displaystyle f(x)=\sum _{i=1}^{n}x_{i}g_{i}(x).} For every i = 1 , … , n , {\displaystyle i=1,\ldots ,n,} {\displaystyle i=1,\ldots ,n,} let α i := g i ( y ) {\displaystyle \alpha _{i}:=g_{i}(y)} {\displaystyle \alpha _{i}:=g_{i}(y)} where 0 = f ( y ) = ∑ i = 1 n y i g i ( y ) = g n ( y ) {\displaystyle 0=f(y)=\sum _{i=1}^{n}y_{i}g_{i}(y)=g_{n}(y)} {\displaystyle 0=f(y)=\sum _{i=1}^{n}y_{i}g_{i}(y)=g_{n}(y)} implies α n = 0. {\displaystyle \alpha _{n}=0.} {\displaystyle \alpha _{n}=0.} Then for any x = ( x 1 , … , x n ) ∈ R n , {\displaystyle x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n},} {\displaystyle x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n},} f ( x ) = ∑ i = 1 n x i g i ( x ) = ∑ i = 1 n [ x i ( g i ( x ) − α i ) ] + ∑ i = 1 n − 1 [ x i α i ]  using  g i ( x ) = ( g i ( x ) − α i ) + α i  and  α n = 0 = [ ∑ i = 1 n x i ( g i ( x ) − α i ) ] + [ ∑ i = 1 n − 1 x i x n α i ] + [ ∑ i = 1 n − 1 x i ( 1 − x n ) α i ]  using  x i = x n x i + x i ( 1 − x 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}}} {\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 3 n − 2 {\displaystyle 3n-2} {\displaystyle 3n-2} terms above has the desired properties. ◼ {\displaystyle \blacksquare } {\displaystyle \blacksquare }

See also

Citations

  1. Nestruev 2020, pp. 17–18.

References