I am combating the next drawback:

Let $fcolonmathbb{R}^2tomathbb{R}$ be a twice constantly

differentiable operate satisfying $$f(0,y)=0mbox{ for

all }yinmathbb{R}$$(a) Present that $f(x,y) = xg(x,y)$ for all pairs $(x,y)inmathbb{R}^2$,

the place $g$ is the operate given by $$g(x,y) =

intlimits_0^1frac{partial f}{partial x}(tx,y)dt$$(b) Present that $g$ is constantly differentiable and that for all

$xinmathbb{R}$ $$g(0,y) = frac{partial f}{partial

x}(0,y),~frac{partial g}{partial y}(0,y) = frac{partial^2

f}{partial xpartial y}(0,y)$$(c) If $frac{partial f}{partial x}(0,0)neq0$ there’s a

neighborhood $V$ of $(0,0)$ in $mathbb{R}^2$ such that $f^{-1}(0)cap

V = Vcap {x=0}$(d) If $frac{partial f}{partial x}(0,0) = 0$ and $frac{partial^2

f}{partial xpartial y}(0,0)neq0$ there’s a neighborhood $V$ of

$(0,0)$ in $mathbb{R}^2$ such that $f^{-1}(0)cap V$ consists of the

union of the set $Vcap{x=0}$ with a curve by way of $(0,0)$, whose

tangent at $(0,0)$ just isn’t vertical (not parallel to the $y$-axis)

Listed here are my makes an attempt:

(a) Observe that $frac{partial f}{partial x}(tx,y) = frac{t}{x}cdotfrac{partial f}{partial t}(tx,y)$ and so utilizing the combination by elements

$$g(x,y) = intlimits_0^1frac{partial f}{partial x}(tx,y)dt = intlimits_0^1 frac{t}{x}cdotfrac{partial f}{partial t}(tx,y)dt = frac{1}{x}left(f(x,y) – intlimits_0^1 f(tx,y)dtright)$$

it suffices to indicate that $intlimits_0^1 f(tx,y)dt = 0$, nonetheless, I am unable to see learn how to proceed with it.

(b) Expressions for $g(0,y)$ and $frac{partial g}{partial y}(0,y)$ is the results of a simple calculation utilizing the definiton of $g(x,y)$, however I’ve troubles displaying that $g$ is constantly differentiable. What am I purported to do — show that the partial derivatives exist and they’re steady? If that’s the case, can somebody write down the small print?

(c) One can repair $y=0$ and think about a operate $f(x,0)colonmathbb{R}tomathbb{R}$. Then, since $frac{partial f}{partial x}(0,0)neq0$, the Inverse Operate Theorem says that there exists a neighborhood $U$ of $0inmathbb{R}$ such that $fcolon Uto f(U)$ is the bijection. Denote $V = Utimes{0}$, so $Vsubsetmathbb{R}^2$ and we’re executed. Am I proper right here?

(d) It looks like utilizing the results of half (b) we are able to do the identical trick with the operate $g(0,y)colonmathbb{R}tomathbb{R}$, however I am unable to end the proof.

Any assist will probably be actually appreciated.Thanks so much prematurely.