# Proof of a property of a function where the total derivative is the same at every point

Say we have a function $f: \mathbb{R}^n\to \mathbb{R}^p$ about which we know that it is totally differentiable, and thus it has some total derivative, and we know it's is the total derivative in any location, in other words: $Df(x)$ is the same for every $x \in \mathbb{R}^n$.

In one dimension this would suggest that the function is a linear function: $f(x) = ax+b$ for some $a$ and $b$ in the reals. Thus we know something else: there exists some real $c$ such that $\forall x\in\mathbb{R}:f(x) = x*f'(x) + c$.

Our professor suggested that this would still remain true in multiple dimensions, in other words: For the function $f: \mathbb{R}^n\to\mathbb{R}^p$ with $Df(x)=A$ where $A$ is the same for every $x\in \mathbb{R}^n$, there is some $c\in\mathbb{R}^p$ such that $f(x)=A(x)+c$ for all $x\in\mathbb{R}^n$.

I don't really know how to prove this. It appeared to me that in the one-dimensional case this would be true by simply using a Taylor series, but I'm not sure how to do that when the function is $\mathbb{R}^n\to\mathbb{R}^p$ (how do we make sure the 'second' derivative would be $0$ such that it would be of the same form as in one dimension?) Also, it was suggested to think about the fact that if $p=1$ and the gradient of $f-A$ is $0$, then you could prove $f-A$ is constant, but I'm not sure how to generalize this to higher values of $p$. Perhaps do it pointwise (prove the truth of this for every element of the vector?)

Can someone help me a bit further? I'm not sure where to go from here.

• Do you know the definition of derivative for dimensions greater than $1$? If you do, this is almost an exercise by definition. – астон вілла олоф мэллбэрг Sep 21 '16 at 9:57
• As far as I'm aware it's some linear tranformation $A$ such that the limit of $x\to y$ over $||f(x)-f(y)-A(x-y)||/||x-y||$ goes to $0$, right? – user370954 Sep 21 '16 at 10:04
• Yes, you are right. Now, can you use this definition, along with the fact that $A(x-y) = C(x-y)$ for some constant matrix $C$? – астон вілла олоф мэллбэрг Sep 21 '16 at 10:05
• I am not sure how I would use that, except maybe we could rewrite the limit to the limit of $x\to y$ of $||f(x)-f(y)||/||x-y|| + c \geq 0$, but I'm not sure if that helps (I just saw your edit after typing this) – user370954 Sep 21 '16 at 10:11
• Or we could possibly rewrite it to: $\lim{x \to y} ||f(x)-C(x)-f(y)+C(y)||/||x-y|| = 0$. Which makes me think (but unable to prove) $||f(x)-C(x)|| = ||f(y)+C(y)||$ which closely resembles (but still unable to prove) that $f(x)-C(x)$ is equal for all $x$ and thus some constant – user370954 Sep 21 '16 at 10:22

The definition of differentiability in $R^n$ is as follows:

$f$ is differentiable at a point $x$ if there exists a linear transformation $A_{fx}$ depending on $f$ and $x$ such that $\frac{||f(x+h)-f(x) -A_{fx}(x)||}{||h||} \to 0$.

Now, consider the function $g(x) = f(x) - Cx$, where $A_{fx} = C$ the constant derivative. Then, note that $\frac{||g(x+h)-g(x)||}{||h||} \to 0$. Now, we can use something called the mean value theorem for many dimensions:

Let $[x,y]$ be the line joining $x$ and $y$ (this is inside $\mathbb{R}^n$ clearly, but can be used when the domain is convex) . Then, there exists $z \in [x,y]$ such that $A_{fz}(x-y) = g(y)-g(x)$.

The proof for this is similar to the mean value theorem for one dimension.

Now, The result is immediate: since the derivative of $g$ is zero everywhere, $g(y)-g(x) = A_{fz}(x-y) = 0$ for all $x$ and $y$. Let $d = g(x)$ be the constant function represented by $g$, then $f(x) = Cx+d$. Hence the result follows.

• I think I get it now, thank you very much! – user370954 Sep 21 '16 at 10:28
• A small apology from my part. I would like to add that this can't be done without the mean value theorem. The reason for that is that differentiation is a "local" property (depends on the neighbourhood of a point) while requiring constancy is a global property (over the entire space). Hence, the introduction of the mean value theorem is used to take advantage of the fact that the derivative at other points is also given as a constant. Without this, you will not be able to get beyond the derivative at the point itself, so you won't be making full use of the hypothesis of the question. – астон вілла олоф мэллбэрг Sep 21 '16 at 10:37