Let $f:\mathbb{R}^n\to\mathbb{R}$ be such that $f(\lambda x)=\lambda f(x)$ for any $x \in \mathbb{R}^n$ and any $\lambda\in\mathbb{R}$.

I'm looking to prove that the partial derivatives of $f$ exist, however I don't have much of an idea on how to begin. I took a guess based on what I can work with and thought that if the partial derivatives $\frac{\partial f}{\partial x_i}$ exist, they would have to satisfy the following equation (1):

$$\frac{\partial f}{\partial x_i}(\lambda x_1,\cdots,\lambda x_i,\cdots,\lambda x_n)=\lambda\frac{\partial f}{\partial x_i}(x_1,\cdots,x_i,\cdots,x_n)$$

so I went and calculated the partials:

$$\lim_{h\to 0}\frac{f(\lambda x_1,\cdots,\lambda (x_i+h),\cdots,\lambda x_n)-f(\lambda x_1,\cdots,\lambda x_i,\cdots,\lambda x_n)}{h}$$

using the hypothesis, we certainly have that the limit above is equal to

$$\lim_{h\to 0}\lambda \cdot\frac{f(x_1,\cdots,(x_i+h),\cdots,x_n)-f(x_1,\cdots,x_i,\cdots,x_n)}{h}$$

and so equation (1) is satisfied. I feel like this assumes existence of partial derivatives, so it's obviously not what I'm trying to prove. But I haven't thought about any other possible approach...

  • $\begingroup$ This is false. This function does not even have to be continuous. $\endgroup$ – Hans Jun 14 '18 at 21:52
  • $\begingroup$ I'm also skeptical of this, if we had the other linearity condition ($f(x+y)=f(x)+f(y)$) it could be easily proved. I'm gonna assume there is missing information on the question. $\endgroup$ – AstlyDichrar Jun 14 '18 at 22:03

We show in general such a function does not have well defined partials at every point of $\mathbb{R}^{n}$.

Take $n = 2$, $f:\mathbb{R^2} \to \mathbb{R}$ given by

$$f(x,y) = \begin{cases} \frac{x^2 + y^2}{x} &\text{ if } x \neq 0 \\ 0 &\text{ if } x = 0. \end{cases}$$

Let $\lambda \in \mathbb{R}$. If $\lambda = 0$ then $f(\lambda x, \lambda y) = f(0,0) = 0 = \lambda f(x,y)$. Consider $\lambda \neq 0$. If $x \neq 0$ then $\lambda x \neq 0$ so $$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{\lambda x} = \frac{\lambda^2}{\lambda}f(x,y) = \lambda f(x,y)$$. If $x = 0$, then $\lambda x = 0$ so $$f(\lambda x, \lambda y) = 0 = \lambda 0 = \lambda f(x,y)$$. Thus for any $\lambda \in \mathbb{R}$ and $(x,y) \in \mathbb{R}^2$, $f(\lambda x, \lambda y) = \lambda f(x,y)$.

Next I claim that $f$ does not have a partial at $(0,1)$ with respect to $x$. We have $$\lim_{h \to 0}\frac{f(0 + h, 1) - f(0,1)}{h} = \lim_{h \to 0} \frac{h^2 + 1}{h} = \lim_{h \to 0} h + \frac{1}{h}\to \infty.$$ Thus in general a function with this property will not have partial derivatives at every point.


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