# If $f$ and $\frac{\partial^2 f}{\partial x^2}$ are continuous, then $\frac{\partial f}{\partial x}$ is continuous?

Let $$f:\mathbb{R}^2\rightarrow\mathbb{R}$$ be a function such that both $$f$$ and $$\frac{\partial^2 f}{\partial x^2}$$ are continuous on $$\mathbb{R}^2$$. Is then $$\frac{\partial f}{\partial x}$$ also continuous on $$\mathbb{R}^2$$?

Precise statement: Let $$g:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto\lim_{\varepsilon\rightarrow0}\frac{f(x+\varepsilon,y)-f(x,y)}{\varepsilon},$$ and let: $$h:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto\lim_{\varepsilon\rightarrow0}\frac{g(x+\varepsilon,y)-g(x,y)}{\varepsilon},$$ then, assuming $$g$$ and $$h$$ both exist, if $$f$$ and $$h$$ are continuous, is $$g$$ also continuous?

This looks quite elementary but unusual as we don't deal with any $$C^k$$ function. I am quite lost with this one. Any help?

• $f_y$ often means $\partial f/\partial y$. Your $df/dx$ is what most would call $\partial f/\partial x$; I recommend this notation Commented Jul 16 at 18:14
• So I'd phrase it as: if $f$ and $\frac{\partial^2f}{\partial x^2}$ are continuous, is $\frac{\partial f}{\partial x}$ continuous? Commented Jul 16 at 18:15
• @FShrike Thank you, I applied your suggestion and further clarified the question. Commented Jul 16 at 18:21
• I see my mistake : ) My deleted answer assumed (and this was very silly) that $\partial f/\partial x$ is continuous along the $y$ direction... at just one point Commented Jul 16 at 18:28
• @sudeep5221 can you please delete your comment? people keep upvoting it and i worry they mistake your comment for a solution, which it is not. i just want to reduce confusion, it is not personal Commented Jul 16 at 20:25

Yes, $$\frac{\partial f}{\partial x}$$ is necessarily continuous. Using $$g$$ and $$h$$ as defined in your question: Let $$H$$ be any second antiderivative of $$h$$ with respect to $$x$$. Integrating $$h$$ twice, we find that $$f$$ admits the form $$H(x, y) + j_{1}(y)x + j_{2}(y) \quad \text{for some} \quad j_{1}, j_{2}.$$ We're given that this function is continuous; we wish to show that $$g = \frac{\partial}{\partial x} H(x, y) + j_{1}(y)$$ is continuous too. This begins the goose chase:
1. $$H(x, y)$$ and $$\frac{\partial}{\partial x} H(x, y)$$ are both continuous (as integrals of the continuous function $$h$$). Hence $$j_{1}(y)x + j_{2}(y)$$ is continuous.
2. $$j_{2}(y)$$ is continuous since $$j_{1}(y)x + j_{2}(y)$$ is continuous along the path $$(0, t)$$. Therefore $$j_{1}(y)x$$ is continuous.
3. $$j_{1}(y)$$ is continuous since $$j_{1}(y)x$$ is continuous along the path $$(1, t)$$.
So $$g$$ is continuous as desired.