# Equality of mixed partial derivatives

Is the following statement $$\frac{\partial^2 f}{\partial x \, \partial y}=\frac{\partial^2 f}{\partial y \, \partial x}$$ always true? If not what are the conditions for this to be true?

• Very nice result here: math.stackexchange.com/a/47885/27978 Commented Dec 20, 2014 at 17:15
• Just a remark: Note that in a broader sense, if you replace the derivatives with covariant derivatives on a Riemannian manifold (concider that everything is happening on a sphere, for example), the difference between $f_{xy}$ and $f_{yx}$ is the curvature (at least when differentiating vector fields). So, you can interpret the above identity as saying that $\mathbb{R}^n$ is "flat". Commented Dec 20, 2014 at 17:21
• Here, you can find the more generalized version of the above theorem. math.stackexchange.com/questions/3439436, mathoverflow.net/questions/346161 Commented Sep 3, 2020 at 4:49
• And here's the standard counterexample with $f_{xy}(0,0) \neq f_{yx}(0,0)$: math.stackexchange.com/questions/219759/… Commented Oct 21, 2021 at 7:20

Second order partial derivatives commute if $f$ is $\mathcal{C}^2$ (i.e. all the second partial derivatives exist and are continuous). This is sometimes called Schwarz's Theorem or Clairaut's Theorem; see here.

This is true in general if $f \in \mathcal{C}^2$. This has a name: symmetry. More formally, it is known as Clariut's Theorem or Schwarz's theorem.

The equality of mixed directional derivatives holds if $$f$$ is twice differentiable at the given point. There a quick easy (and nice) proof using Fubini's theorem - but it assumes that $$f\in C^2$$. It is a striking fact (IMHO) that the symmetry already follows from point-wise differentiability. I've learned this from the (excellent) "Advanced calculus" book by Loomis and Sternberg (an electronic copy is freely available at https://people.math.harvard.edu/~shlomo), theorem 16.3 (page 187).

I shall reproduce the proof below, correcting a small error (as I understand it) that confused me a quite a bit.

We first recall the definition: Let $$V,W$$ be normed linear spaces (over $$\mathbb R$$). A function $$f:V\longrightarrow W$$, defined at some neighborhood of $$v_0\in V$$, is said to be differentiable at $$v_0$$ if there exists $$df|_{v_0}\in Hom(V,W)$$ (necessarily unique) such that $$\lim_{v\to 0_V}\frac{f(v_0+v)-f(v_0) - df|_{v_0}(v)}{\|v\|} = 0_W$$ The uniqueness follows from the observation that the zero map is the only $$S\in Hom(V,W)$$ such that $$\lim_{v\to 0_V}\frac{S(v)}{\|v\|} = 0_W$$.

Recall that for any $$v\in V$$ the directional derivative is defined as $$\frac{\partial f}{\partial v}(v_0)=\lim_{t\to 0}\frac{f(v_0+tv)-f(v_0)}{t}$$ assuming the limit exists. It is easy to see that if $$df|_{v_0}$$ exists than the directional derivatives exist and $$\frac{\partial f}{\partial v}(v_0)=df|_{v_0}(v)$$ Proof: Plug in $$tv$$ for $$v$$ in the definition of to get $$\frac{f(v_0+tv)-f(v_0) - df|_{v_0}(tv)}{\|tv\|} = \frac{1}{\|v\|}\cdot\Big(\frac{f(v_0+tv)-f(v_0) }{t} - df|_{v_0}(v)\Big)$$ and this by assumption tends to $$0_W$$ as $$t\to 0$$. $$\$$q.e.d.
Assuming that $$f$$ is differentiable, it follows that the map $$\frac{\partial f}{\partial v}:V\longrightarrow W$$ is equal to the composition $$V\overset{df|_{(\cdot)}}{\longrightarrow} Hom(V,W) \overset{eval_v}{\longrightarrow}W\$$ where $$eval_v$$ is the (linear) map $$T\mapsto T(v)$$

If $$f$$ is differentiable on some neighborhood of $$v_0$$, we can ask whether the map $$df|_{(\cdot)}:V\longrightarrow Hom(V,W)$$ is itself differentiable at $$v_0$$. If so then $$d^2f|_{v_0}:=d\big(df|_{(\cdot)}\big)|_{v_0}$$ is an element of $$Hom\big(V,Hom(V,W)\big)$$. Also : $$\frac{\partial^2 f}{\partial u \partial v}(v_0)= \frac{\partial\big(\frac{\partial f}{\partial v}\big)}{\partial u}(v_0)= d\Big(\frac{\partial f}{\partial v}\Big)|_{v_0}(u)= d\big(eval_v\circ df|_{(\cdot)}\big)|_{v_0}(u)= \big(eval_v\circ d^2f|_{v_0}\big)(u)= eval_v\big(d^2f|_{v_0}(u)\big)= \big(d^2f|_{v_0}(u)\big)(v)$$

Now we can finally state the theorem: If $$f$$ is twice differentiable at $$v_0$$ then for all $$u,v\in V\$$ $$\frac{\partial^2 f}{\partial u \partial v}(v_0) =\frac{\partial^2 f}{\partial v \partial u}(v_0)$$ Proof: We need to show that the map $$\big(d^2f|_{v_0}(u)\big)(v):V\times V\longrightarrow W$$ is symmetric.
Fix some $$\epsilon > 0$$.
By the definition of $$d^2f|_{v_0}$$, $$\ \exists\delta>0$$ such that $$\forall x\in V$$ $$\|x\|<\delta\ \Rightarrow\ \big\| df|_{v_0+x}-df|_{v_0} - d^2f|_{v_0}(x) \big\|_{Hom(V,W)} <\epsilon\cdot\|x\|$$ Apply this also to some (small) $$x+y$$ and subtract to conclude that $$\forall x,y\in V$$ $$\Big(\|x\|<\delta\ \& \ \|x+y\|<\delta \Big)\ \Rightarrow\ \big\| df|_{v_0+x}-df|_{v_0+x+y} - d^2f|_{v_0}(-y) \big\| <\epsilon\cdot\big(\|x\| + \|x+y\|\big)$$ Now fix some (non-zero) $$y\in V$$ such that $$0<\|y\|<\delta/3$$.
Clearly $$\|x\|\leq 2\|y\| \Rightarrow\ \|x\|<\delta\ \& \ \|x+y\|<\delta \ \& \ \big(\|x\| + \|x+y\|\big) \leq 5\|y\|$$
So setting $$g(x):=f(v_0+x)-f(v_0+x+y)-d^2f|_{v_0}(-y)(x)$$ we have that $$\forall x\in B_{2\|y\|} \ \| dg|_x \| < 5 \epsilon\cdot \|y\|$$ (I'm using $$B_a := \{x\in V:\|x\|\leq a\})$$).
Therefore (from the mean value theorem) $$\forall a,b\in B_{2\|y\|}\quad \|g(a)-g(b)\|_W < 5 \epsilon\cdot \|y\|_V \cdot \|a-b\|_V$$ i.e. $$\big\| f(v_0+a) - f(v_0+a+y) - f(v_0+b) + f(v_0+b+y) - d^2f|_{v_0}(-y)(a-b) \big\| < 5 \epsilon\cdot \|y\| \cdot \|a-b\|$$ For any $$x\in B_{\|y\|}$$ we can apply the above with $$a=-y,b=-y-x$$ (note that indeed $$a,b\in B_{2\|y\|}$$) and we get that $$\big\| f(v_0-y) - f(v_0) - f(v_0-y-x) + f(v_0-x) - d^2f|_{v_0}(-y)(x) \big\| < 5 \epsilon\cdot \|y\| \cdot \|x\|$$ Let's recap: Denoting $$S(x,y) = f(v_0) + f(v_0-y-x) - f(v_0-y) - f(v_0-x)$$ we have that for any $$0<\|y\|<\delta/3$$ and $$\|x\|\leq \|y\|$$ it holds that $$\big\| d^2f|_{v_0}(y)(x) - S(x,y) \big\| < 5 \epsilon\cdot \|y\| \cdot \|x\|$$ Clearly $$S(x,y)=S(y,x)$$ and therefore combined with the same inequality with reversed order we conclude that $$0< \|x\| = \|y\|<\delta/3 \Rightarrow \big\| d^2f|_{v_0}(y)(x) - d^2f|_{v_0}(x)(y) \big\| < 10 \epsilon\cdot \|y\| \cdot \|x\|$$ Using the homogeneity of $$d^2f|_{v_0}(\cdot)(\cdot)$$, it follows that $$\|x\| = \|y\| = 1 \Rightarrow \big\| d^2f|_{v_0}(y)(x) - d^2f|_{v_0}(x)(y) \big\| \leq 10 \epsilon\cdot \delta^2/9$$ (apply the previous inequality to $$tx,ty$$ with $$t<\delta/3$$)
Since $$\epsilon$$ was arbitrary (and we may assume $$\delta<1$$), it follows that $$d^2f|_{v_0}(y)(x)=d^2f|_{v_0}(x)(y)$$ for any pair of unit vectors. Applying homogeneity once more we see that it holds without restriction.
q.e.d.