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Let $f:\textbf{R}^{2}\to\textbf{R}$ be the function defined by $f(x,y) := \frac{x^{3}}{x^{2}+y^{2}}$ when $(x,y)\neq(0,0)$, and $f(0,0) = 0$. Show that $f$ is not differentiable at $(0,0)$, despite being differentiable in every direction $v\in\textbf{R}^{2}$ at $(0,0)$.

MY ATTEMPT

Let us prove the directional derivative part first. Given $v = (a,b)\neq(0,0)$, we have that \begin{align*} D_{v}f(0,0) = \lim_{t\to 0^{+}}\frac{f((0,0) + t(a,b)) - f(0,0)}{t} = \lim_{t\to 0^{+}}\frac{f(ta,tb)}{t} = \lim_{t\to 0^{+}}\frac{t^{3}a^{3}}{t^{3}(a^{2}+b^{2})} = \frac{a^{3}}{a^{2}+b^{2}} \end{align*}

Hence $f$ is differentiable in every direction $v\in\textbf{R}^{2}$ at $(0,0)$.

In particular, we have that \begin{align*} \frac{\partial f}{\partial x}(0,0) = D_{e_{1}}f(0,0) = 1\quad\wedge\quad\frac{\partial f}{\partial y}(0,0) = D_{e_{2}}f(0,0) = 0 \end{align*}

Now it remains to prove that $f$ is not differentiable.

Suppose otherwise that $f$ is differentiable at $(0,0)$. Then we would have \begin{align*} f'((0,0)) = \left(\frac{\partial f}{\partial x}(0,0), \frac{\partial f}{\partial y}(0,0)\right) = (1,0) \end{align*}

Similarly, we should have \begin{align*} \lim_{x\to(0,0);x\neq(0,0)}\frac{\|f(x,y) - f(0,0) - (1,0)((x,y) - (0,0))\|}{\|(x,y) - (0,0)\|} = \lim_{x\to(0,0);x\neq(0,0)}\left|\frac{xy^{2}}{(x^{2}+y^{2})^{3/2}}\right| = 0 \end{align*}

But this is not the case. If it was the case, the following limits should be equal due to the property of composition of limits: \begin{align*} \lim_{t\to 0}f(\Gamma_{1}(t)) = \lim_{t\to 0}f(\Gamma_{2}(t)) \end{align*} whenever $\Gamma_{1}(t)\to(0,0)$ and $\Gamma_{2}(t)\to(0,0)$ as $t\to 0$.

Having said that, it suffices to consider the curves $\Gamma_{1}(t) = (t,t)$ and $\Gamma_{2}(t) = (2t,t)$, whence we get \begin{align*} \lim_{t\to 0}f(\Gamma_{1}(t)) = \lim_{t\to 0}\left|\frac{t^{3}}{(t^{2}+t^{2})^{3/2}}\right| = \lim_{t\to 0}\frac{1}{2\sqrt{2}} = \frac{1}{2\sqrt{2}} \end{align*} as well as \begin{align*} \lim_{t\to 0}f(\Gamma_{2}(t)) = \lim_{t\to 0}\left|\frac{2t^{3}}{(4t^{2}+t^{2})^{3/2}}\right| = \lim_{t\to 0}\frac{2}{5\sqrt{5}} = \frac{2}{5\sqrt{5}} \end{align*}

which contradicts our assumption, and we are done.

Could someone please verify if the wording of my proof is formal enough? Am I missing any step?

user0102
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2 Answers2

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It looks good to me. Just a few comments and an alternative easier solution.

If it was the case, the following limits should be equal due to the property of composition of limits: \begin{align*} \lim_{t\to 0}f(\Gamma_{1}(t)) = \lim_{t\to 0}f(\Gamma_{2}(t)) \end{align*} whenever $\Gamma_{1}(t)\to(0,0)$ and $\Gamma_{2}(t)\to(0,0)$ as $t\to 0$. Having said that, it suffices to consider the curves $\Gamma_{1}(t) = (t,t)$ and $\Gamma_{2}(t) = (2t,t)$

This is to much in my opinion. The person reading your paper on differentiability is surely familiar with the basic properties of limits. You can juste replace it with something like "but taking the limit along $x= y$ and $2x = y$ gives different results ..."

For an alternative (shorter) proof that it is not differentiable at $(0,0).$

If $f$ was differentiable at $v = (a,b)$ then the directional derivative at $(0,0), $ $D_vf(0,0),$ exists in every direction and

$$ D_vf(0,0) = \nabla f(0) \cdot v = a\frac{\partial f}{\partial x}(0) + b\frac{\partial f}{\partial y}(0)=a$$ but we know $D_vf(0) = \frac{a^3}{a^2 + b^2}$ this is a contradiction.

If $f$ is differentiable then the directional derivative of $D_vf$ at any point must be linear in $v !$

Digitallis
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Your answer is absolutely what is expected. I would just correct a minor detail. The proper way to write the differentiability condition is as follows:

\begin{align*} \lim_{h\to0}\frac{\|f(x_0+h) - f(x_0) - J(h)\|}{\|h\|} = 0 \end{align*} so: \begin{align*} \lim_{(x,y)\to(0,0)}\frac{\|f((0,0)+(x,y)) - f(0,0) - (1,0)\cdot(x,y)\|}{\|(x,y)\|} = 0 \end{align*}

which gives you the same result in this case.

Bcpicao
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