I got stuck on a fairly simple exercise in differential calculus of multiple variables. I shall present the exercise in it's entierty to provide some context:
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be partially differentiable continuously in all of $\mathbb{R}^2$. It is also given that: \begin{gather*} \forall p \in \mathbb{R}^2 \\ (*) \quad \frac{\partial f}{\partial x}(p) = \frac{\partial f}{\partial y}(p) \\ \end{gather*}
In addition, it is given that for some $\alpha$, $\frac{\partial f}{\partial x}(\alpha) \geq 0$ and that $\gamma = |\nabla f(\alpha)|$, so it is easily obtainable that $\nabla f(\alpha) = (\frac{\sqrt{2}}{2}\gamma, \frac{\sqrt{2}}{2}\gamma)$ which is essentially what you were required to show in the first part of the question.
My misunderstandings begin with the second part, which is certainly the more challenging one, and is phrased thus:
For all $(x,y) \in \mathbb{R}^2$ show that $f(x,y) = f(x + y, 0)$
Now, this is completely understandable intuitively by looking at $(*)$ - roughly means that what ever may be the path we were to take, travelling in the $x$ direction will always have the same effect as travelling in the $y$ direction.
I am having trouble formalizing a proof... I wanted to use the chain rule and my professor assured me this is a valid way but -
- What exactly should I derive? I think the following could lead somewhere: $\nabla(f(x, y) - f(x + y, 0))$
- I am not sure how to derive the function in the right side of the equality using the chain-rule (this is a technique problem)
- Even if it turns out that the gradients of both terms are equal, it doesn't necessarily mean that the functions are equal in all of $\mathbb{R}^2$, if I am not mistaken, so I am not quite sure what's the next step.
Many thanks for reading my question, hopefully it was phrased soundly enough.
Any suggestions or directions would be extremely appreciated. Thanks again!
The following is a formalization of idea suggested by @ancientmathematician, thanks to his extremely helpful comments above.
Let $s = x + y$ for some $(x,y) \in \mathbb{R}^2$. Let us define:
\begin{gather*} F: \mathbb{R} \rightarrow \mathbb{R} \\ \lambda \rightarrow f(\lambda s, (1 - \lambda)s) \\ \\ \varphi: \mathbb{R} \rightarrow \mathbb{R}^2 \\ \lambda \rightarrow (\lambda s, (1 - \lambda)s) \end{gather*}
Of course, $F$ is differentiable in $\mathbb{R}$ by the chain rule, and by what we know about $f$.
Let $a,b \in \mathbb{R}$, by Lagrange's Mean Value Theorem we know $\exists \xi \in (a, b)$ s.t:
\begin{gather*} F(a) - F(b) = \frac{dF}{d\lambda}(\xi)\cdot(a - b) \end{gather*}
Let us evaluate $\dfrac{dF}{d\lambda}$ in an arbitrary point using the Chain Rule: \begin{align*} & \frac{dF}{d \lambda}(\lambda) = \nabla f(\lambda s, (1-\lambda)s) \cdot \nabla \varphi (\lambda) \\ & = \Big(\frac{\partial f}{\partial x}(\lambda s, (1-\lambda)s), \frac{\partial f}{\partial y}(\lambda s, (1-\lambda)s)\Big) \cdot (s, -s) \end{align*}
Now, given the hypothesis that $\forall \alpha \in \mathbb{R}^2 : \dfrac{\partial f}{\partial x}(\alpha) = \dfrac{\partial f}{\partial y}(\alpha)$, the last term yields $0$ when the product is evaluated. So we can claim $F(a) - F(b) = \frac{dF}{d\lambda}(\xi)\cdot(a - b) = 0$.
This is true $\forall a,b \in \mathbb{R}$, and therefore true for $a = \frac{x}{s}$ and $b = 1$. This specific case yields the following expression: \begin{gather*} F(\frac{x}{s}) - F(1) = f(x, y) - f(x + y, 0) = 0 \end{gather*}
So, finally we have $\forall (x,y) \in \mathbb{R}^2 : f(x,y) = f(x+y, 0)$.