Exercise 4.7

Question

Exercise 7:
    If $E\subset x$ and if $f$ is a function defined on $X$, the {\it restriction} of $f$ to $E$ is the function $g$ whose domain of definition is $E$, such that $g(p)=f(p)$
    for $p\in E$. Define $f$ and $g$ on $\mathbf R^2$ by: $f(0,0)=g(0,0)=0$, $$f(x,y)=\frac{xy^2}{x^2+y^4},\quad g(x,y)=\frac{xy^2}{x^2+y^6}$$ if $(x,y)
eq(0,0)$. Prove that
    $f$ is bounded on $\mathbf R^2$, that $g$ is unbounded in every neighborhood of $(0,0)$, and that $f$ is not continuous at $(0,0)$; nevertheless, the restrictions of both
    $f$ and $g$ to every straight line in $\mathbf R^2$ are continuous!

analambanomenos · Accepted Answer

Note that both $f$ and $g$ are equal to 0 on the $x$ and $y$ axes.

For $k\in\mathbf R$, $f(ky^2,y)=k/(k^2+1)$ is constant for $y
eq0$. Since the value of $f$ along the parabola $(ky^2,y)$ drops from $k/(k^2+1)$ to 0 at $y=0$, $f$ is not
    continuous at $(0,0)$. These parabolas sweep out $\mathbf R^2$ except for the $x$ axis, and the values of $f$ along these parabolas reach a maximum value of $1/2$ for $k=1$,
    so the values of $f$ lie in $[0,1/2]$.

For $y
eq 0$, $g(ky^3,y)=k/\bigl((k^3+1)y\bigr)ightarrow\infty$ as $yightarrow 0$, so $g$ is unbounded in every neighborhood of $(0,0)$.

Since $f$ and $g$ are continuous away from the origin, their restrictions to any line which doesn't intersect the origin is also continuous. And since $$f(x,kx)=
    \frac{k^2x}{1+k^4x^2}\quad\quad g(x,kx)=\frac{k^2x}{1+k^6x^4}$$ restrictions of $f$ and $g$ to lines which go through the origin are also continuous.

Exercise 4.7

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Exercise 4.7

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