Let $g\colon\mathbb{R} \to \mathbb{R}$, and define $f\colon\mathbb{R}^2 \to \mathbb{R}$ by $f(x,y)=yg(x)$.

Show that $f$ is differentiable at $(0,0)$ if $g$ is continuous at $0$.

So $f$ is differentiable at $(0,0)$ if $f$ can be written as $$ f(x,y)=f(0,0)+L(x,y)+R(x,y) $$ where $L$ is a linear map consisting out of the partial derivatives which are $0$ with respect to $x$ and $g(0)$ in respect to $y$.

How can I show that $R(x,y)$ goes to $0$. I'm having trouble to conclude the right limit.

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Jun 16 '17 at 15:46
  • $\begingroup$ First guess the limit. What would ${\partial f(0,0) \over \partial x}, {\partial f(0,0) \over \partial y}$ be? $\endgroup$
    – copper.hat
    Jun 16 '17 at 15:52
  • $\begingroup$ ∂f(0,0)/∂x=o ∂f(0,0)/∂y=g(0). Well the limit must be zero anyway, for my statement to be true but i dont understand how i need to built the limit of R(x,y) in the end i want something like R(x,y) has a limit at 0 if g(0) is continous in 0. sorry for not using mathjax $\endgroup$
    – johnka
    Jun 16 '17 at 16:02
  • $\begingroup$ Well, you have guessed the limit correctly, what are the remaining issues? $\endgroup$
    – copper.hat
    Jun 16 '17 at 16:27
  • $\begingroup$ $\lim _{ (x,y)\rightarrow (0,0) }{ \frac { R(x,y) }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } =0 } $ i cant see how R(x,y) should look like (i know that this is a rather dumb issue) $\endgroup$
    – johnka
    Jun 16 '17 at 16:36

Once you have identified what $L(x,y)$ should look like, i.e. $$ L(x,y) = \frac{\partial f}{\partial x}(0,0) x+ \frac{\partial f}{\partial y}(0,0)y = 0\cdot x + g(0)y = g(0)y $$ then the only "sensible" choice for $R(x,y)$ is what remains: $$ R(x,y) \stackrel{\rm def}{=} f(x,y) - f(0,0) - L(x,y) = yg(x) - 0 - g(0)y = y(g(x)-g(0). $$

So we need to show that this satisfies what it should, that is $$ \lim_{(x,y)\to (0,0)} \frac{\lvert R(x,y)\rvert}{\lVert (x,y)\rVert_2} = 0 \tag{1}. $$

To do so, fix any $\varepsilon > 0$, and let $\delta > 0$ be (by continuity of $g$ at $0$) such that $\lvert g(x)-g(0)\rvert \leq \varepsilon$ whenever $\lvert x\rvert \leq \delta$.

Then, for any $(x,y)$ such that $\lVert (x,y)\rVert_2 = \sqrt{x^2+y^2} \leq \delta$, we get that $\lvert x\rvert \leq \sqrt{x^2+y^2} \leq \delta$, and thus $$ \frac{\lvert R(x,y)\rvert}{\lVert (x,y)\rVert_2} = \frac{\lvert y\rvert\cdot \lvert g(x)-g(0)\rvert}{\sqrt{x^2+y^2}} \leq \varepsilon \cdot \frac{\lvert y\rvert}{\sqrt{x^2+y^2}} \leq \varepsilon $$ the last inequality as $\lvert y\rvert = \sqrt{y^2} \leq \sqrt{x^2+y^2}$. Since $\varepsilon>0$ was arbitrary, this shows (1), and therefore that $f$ is differentiable at $(0,0)$.


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