Exercise 9.15

Question

Exercise 15: Define $f(0,0)=0$, and put
    $$
        f(x,y)=x^2+y^2-2x^2y-\frac{4x^6y^2}{(x^4+y^2)^2}\qquad\hbox{if }(x,y)\ne(0,0).
    $$
    (a) Prove, for all $(x,y)\in\R2$, that 
    $$
        4x^4y^2\le(x^4+y^2)^2.    
    $$
    Conclude that $f$ is continuous.

(b) For $0\le\theta\le2\pi$, $-\infty<t<\infty$, define
    $$
        g_\theta(t)=f(t\cos\theta,t\sin\theta).
    $$
    Show that $g_\theta(0)=0$, $g_\theta'(0)=0$, $g_{\theta}''(0)=2$. Each $g_\theta$ has therefore a strict local minimum at $t=0$. In other words, the restriction of $f$ to each line through
    $(0,0)$ has a strict local minimum at $(0,0)$.

(c) Show that $(0,0)$ is nevertheless not a local minimum for $f$, since $f(x,x^2)=-x^4$.

analambanomenos · Accepted Answer

(a) We have
    $$
        (x^4+y^2)^2-4x^4y^2 = x^8-2x^4y^2+y^4-4x^4y^2 = x^8-2x^4y^2+y^4=(x^4-y^2)^2 \ge 0
    $$
    so that $4x^4y^2\le(x^4+y^2)^2$. To show that $f$ is continuous, it suffices to show this at $(0,0)$, since $f$ is differentiable otherwise. We have
    $$
        \big|f(x,y)\big| \le x^2+y^2+2x^2|y|+x^2\frac{4x^4y^2}{(x^4+y^2)^2} \le 2x^2+y^2+2x^2|y|
    $$
    which goes to $f(x,y)=0$ as $(x,y)\rightarrow 0$.

(b) We have $g_{\theta}(0)=f(0,0)=0$, and for $t\ne 0$,
    \begin{align*}
        g_{\theta}(t) &= t^2\cos^2\theta + t^2\sin^2\theta-2t^3\cos^2\theta\sin\theta - \frac{4t^8\cos^6\theta\sin^2\theta}{(t^4\cos^4\theta+t^2\sin^2\theta)^2} \
        &= t^2 - 2t^3\cos^2\theta\sin\theta - \frac{4t^4\cos^6\theta\sin^2\theta}{(t^2\cos^4\theta+\sin^2\theta)^2}
    \end{align*}
    The denominator in the last term is nonzero unless $\theta=0$ and $t=0$. Excluding the case $\theta=0$ this expression for $g_{\theta}(t)$ is true for all values of $t$ including 0,
    so we can differentiate it using the usual rules of calculus to get
    \begin{align*}
        g_{\theta}'(t) &= 2t - 6t^2\cos^2\theta\sin\theta - \frac{16t^3\cos^6\theta\sin^4\theta}{(t^2\cos^2\theta+\sin^2\theta)^3} \
        g_{\theta}''(t) &= 2 - 12t\cos^2\theta\sin\theta + \frac{48t^2\cos^6\theta\sin^4\theta(t^2\cos^2\theta-\sin^2\theta)}{(t^2\cos^2\theta+\sin^2\theta)^4}
    \end{align*}
    Hence, for $\theta\ne 0$ we have $g_{\theta}'(0)=0$ and $g_{\theta}''(0)=2$. For the special case $\theta=0$, the formula for $g_{\theta}(t)$ simplifies to $g_0(t)=t^2$, so that we also have
    $g_0'(0)=0$ and $g_0''(0)=2$.

(c) We have
    $$
        f(x,x^2) = x^2+x^4-2x^2x^2-\frac{4x^6x^4}{(x^4+x^4)^2} = x^2+x^4-2x^4-x^2 = -x^4.
    $$

Exercise 9.15

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Exercise 9.15

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