Let f will be function $C^{1}$ in near $x^{*} \in \mathbb{R}^{N}$ then $ f(x) = f(x^{*}) + \int_0^1\nabla{f(x^*+ te)^{T}e \thinspace dt } $

Question

Let f will be function $C^{1}$ in near $x^{*} \in \mathbb{R}^{N}$ then $$ f(x) = f(x^{*}) + \int_0^1\nabla{f(x^*+ te)^{T}e \thinspace dt } $$

Proof

We use theory of Newtona-Leibnitza integrate: $$g(b)- g(a) = \int_a^b g^{'}(x) dx$$ consider a new function: $$\gamma: [0,1] \mapsto \mathbb{R^{N}}$$

$$\gamma(t) = x^{*} + et :[0,1]\mapsto R^{n} $$ and we composition with $f \circ \gamma :[0,1] \mapsto \mathbb{R^{N}} $

$$ f(\gamma(1))- f(\gamma(0)) = \int_0^1 (f \circ \gamma)^{'} (t) dt $$ $$ f(x^{*}+e)- f(x^{*}) = \int_0^1 f(\gamma(t))^{'} \gamma^{'}(t) dt $$ $$ f(x^{*}+e) = f(x^{*}) + \int_0^1 f(\gamma(t))^{'} \gamma^{'}(t) dt \color{red}{\text{how to prove that: ? } \iff } $$

$$ f(x^{*}+e) = f(x^{*}) + \int_0^1 \nabla f(x^{*} + et)e dt $$ because $\gamma^{'}(t) = e$

Answer 1

Let $\phi(t) = f(x^*+t(x-x^*))$.

We have $\phi(1) = \phi(0)+ \int_0^1 \phi'(t)dt$.

We have $\phi'(t) = \langle \nabla f(x^*+t(x-x^*)), x-x^* \rangle$.

Hence $f(x) = f(x^*) + \int_0^1 \langle \nabla f(x^*+t(x-x^*)), x-x^* \rangle dt = \langle \int_0^1 \nabla f(x^*+t(x-x^*)) dt, x-x^* \rangle$.