# Show function $g$ is continuously differentiable

Let $$f\in C^2(\mathbb{R^2})$$ with $$f(0, λ) = 0$$ $$\forall \lambda \in \mathbb{R}$$. Show $$g(x,\lambda):=\begin{cases} \frac{f(x,\lambda)}{x}& \text{ if } x\ne0 \\ \partial_x f(0,\lambda)&\text{ if } x=0 \end{cases}$$ is continuously differentiable.

So I need to show that $$g$$ is continuous and its derivative is also a continuous function. Since $$f\in C^2(\mathbb{R^2})$$ and $$\frac{1}{x}$$ is continous on $$\mathbb{R}\backslash\{0\}$$, $$g(x,\lambda)$$ is continous as a composition/product of continuous functions in both cases and I only need to analyse the point $$x=0$$.

Since $$f(0, λ) = 0$$ I can rewrite the first case to $$\frac{f(x,\lambda) - f(0,\lambda)}{x-0}$$ which is the difference quotient in $$0$$ and one gets $$lim_{x\to0}\frac{f(x,\lambda) - f(0,\lambda)}{x-0} = \partial_xf(0,\lambda),\text{ so } g \text{ is continuous}.$$

Problem: I also need to show the continuity of its derivative: $$\partial_xg(x,\lambda)=\begin{cases} \frac{\partial_xf(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^2}& \text{ if } x\ne0 \\ \partial_{xx} f(0,\lambda)&\text{ if } x=0 \end{cases} \text{ and } \partial_{\lambda}g(x,\lambda)=\begin{cases} \frac{\partial_{\lambda}f(x,\lambda)}{x}& \text{ if } x\ne0 \\ \partial_{\lambda}(\partial_x f(0,\lambda))&\text{ if } x=0 \end{cases}$$

$$\partial_{\lambda}g(x,\lambda)$$ should be the same argument as the continuity for $$g$$ just with a $$\partial_{\lambda}$$ in front of it since the cases of $$g$$ only depend on $$x$$.

Since $$\partial_xf(0,\lambda)=lim_{x\to 0}\frac{f(x,\lambda)-f(0,\lambda)}{x}$$ and $$f(0,\lambda)=0$$ $$\forall \lambda$$, it is $$\partial_{xx}f(0,\lambda)=lim_{x\to 0}\frac{\partial_xf(x,\lambda)-\partial_xf(0,\lambda)}{x}=lim_{x\to 0}\frac{\partial_xf(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^2},$$ so the derivative of $$g$$ is also continuous.

Is my proof correct? Thanks!

• Isn't it the definition of $g(x,\lambda)$ in $x=0$? It is $g(0,\lambda)=\partial_xf(0,\lambda)$ since $x=0$ assumed in this case, so $\partial_xg(0,\lambda)=\partial_x(\partial_xf(0,\lambda))=\partial_{xx}f(0,\lambda)$ in my notations ($\partial_{xx}=\partial_x\partial_x$).
– Uhmm
Commented Jul 6, 2022 at 10:44
• @Al.G. do you mean $\partial_xg(0,\lambda):=lim_{x\to 0}\frac{g(x,\lambda)-g(0,\lambda)}{x}=lim_{x\to 0}(\frac{f(x,\lambda)}{x^2}-\frac{\partial_xf(0,\lambda)}{x})=lim_{x\to 0}\partial_xg(x,\lambda)$?
– Uhmm
Commented Jul 6, 2022 at 11:00
• @Al.G. isn't it enough because for $x\neq 0$ it is $\partial_xg(x,\lambda)=\partial_x\frac{f(x,\lambda)}{x}=\frac{\partial_xf(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^2}$ by quotient rule?
– Uhmm
Commented Jul 6, 2022 at 11:11
• @Al.G. what is the problem by saying $\partial_xg(0,\lambda)=\partial_x(\partial_xf(x,\lambda))$ since it's just the definition of $g$ and $f\in C^2(\mathbb{R}^2)$?
– Uhmm
Commented Jul 6, 2022 at 11:16
• @Al.G. ah yes, now I understand. Thank you!
– Uhmm
Commented Jul 6, 2022 at 11:24

In writing out the cases for $$\partial_{x}g(0,\lambda)$$, you seem to have derived $$\partial_{x}g(0,\lambda)=\partial_{xx}f(0,\lambda)$$ from $$g(0,\lambda)=\partial_{x}f(0,\lambda)$$. But that is not the reason the equality above holds.

Let me show a counterexample where that reasoning fails. Define a function $$h(x):=\begin{cases} 0 & x\neq0\\ f'(0) & x=0 \end{cases}$$ and take $$f(x)=\cos x$$ (it took me a while to come up with this example! :D).

Note that $$h(0)=\cos'(0)=-\sin(0)=0$$ so that $$h \equiv 0$$. Hence, both $$h$$ and $$f$$ are even $$C^\infty$$, i.e. infinitely differentiable functions.

But it is not true that $$h'(0)=f''(0)$$ since $$h'(0)=0$$ but $$f''(0)=-\cos(0)=-1$$!

In your proof, you need to use the way $$g(x)$$ is defined for $$x\neq 0$$. It is important as I hope I made it clear with my counterexample.

Now about the specific problem in question. To show continuity of the derivative $$\partial_{x}$$, you need to compute $$\partial_{x}g(0,\lambda):=\lim_{x\to0}\frac{g(x,\lambda)-g(0,\lambda)}{x}$$ explicitly and show that it equals the limit $$\lim_{x\to0}\left(\partial_{x}g(x,0)\right)$$, and from your calculations you know $$\lim_{x\to0}\left(\partial_{x}g(x,0)\right)=\lim_{x\to0}\left(\frac{\partial_{x}f(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^{2}}\right).$$

To finish, you need to show $$\partial_{x}g(0,\lambda)=\lim_{x\to0}\partial_{x}g(x,\lambda)$$, i.e. the following equality: $$\lim_{x\to0}\frac{g(x,\lambda)-g(0,\lambda)}{x}=\lim_{x\to0}\left(\frac{\partial_{x}f(x,\lambda)}{x}-\frac{f(x,\lambda)}{x^{2}}\right).$$

Which is trivial after you replace $$g$$ on the left with its values by definition.

• Thank you for your help! :)
– Uhmm
Commented Jul 6, 2022 at 11:44
• I updated my answer to include what exactly you are expected to show in the solution (instead of only why yours was incomplete). It turns out what's left is kind of trivial, but anyway, needs to be mentioned. Commented Jul 6, 2022 at 11:57
• Anyway, glad to help :) You can mark my answer as 'Accepted' if you think it solves your problem. Commented Jul 6, 2022 at 11:59