Function positively homogeneous and $k$-linear forms.

Question

Let $f: \mathbb{R}^{m} \to \mathbb{R}$ positively homegeneous of degree $k \in \mathbb{N}$ and $f \in C^{k}$. Then there exists a $k$-linear form $\varphi$ in $\mathbb{R}^{m}$ such that $f(x) = \varphi(x,...,x)$ for all $x \in \mathbb{R}^{m}$.

I know that:

$$f(tx) = t^{k}f(x),$$

and

$$\sum \frac{\partial f}{\partial x_{i}}x_{i} = kf(x) \quad (\textit{Euler relation}),$$

If $\varphi$ is a $k$-linear form, so $$\varphi(x_{1},...,x_{k}) = Cx_{1}\cdot ... \cdot x_{k}.$$

But, I don't know how I can use this informations. I appreciate any hints.

Answer 1

Thus is not true in general. The simplest counterexample is the case $m=1$ and $f(x):=\vert x\vert^k$. For arbitrary $m$ you may do the same by using any norm on $\mathbb{R}^m$ instead of the absolute value.

Let now $f$ be in $C^k$ with $k\geq 1$. (Reaction on the remark below.)
Consider the function $\mathbb{R}^n\times[0,\infty)\ni (x,t)\mapsto f(t\cdot x):=h(x,t)$. Then, using the chainrule, $\frac{\partial h}{\partial x_{i}}(x,t)=\frac{\partial f}{\partial x_{i}}(t\cdot x) t$. Since $h(x,t)=t^k f(x)$ we also have $\frac{\partial h}{\partial x_{i}}(x,t)=t^k \frac{\partial f}{\partial x_{i}}(x)$. Thus $\frac{\partial f}{\partial x_{i}}(t\cdot x)=t^{k-1} \frac{\partial f}{\partial x_{i}}(x)$ for $t>0$ and by continuity also for $t=0$. So (induction) $\frac{\partial f}{\partial x_{i}}(x)=P_i(x)$ where $P_i$ is a homogeneous polynomial of degree $k-1$.
Let $g(t):=f(tx)$. Then $g'(t)=\sum_{i=1}^n\frac{\partial f}{\partial x_{i}}(tx) x_i =\sum_{i=0}^n P_i(tx)x_i=t^{k-1}\sum_{i=0}^n P_i(x)x_i=t^{k-1} P(x)$ where $P$ is a homogeneous polynomial of degree $k$.
Now $g(1)=f(x), g(0)=0$ and $f(x)=g(1)-g(0)=\int_0^1 g'(t)=P(x)\frac{t^k}{k}{\Large\vert}_0^1=\frac{P(x)}{k}$, a homogeneous polynomial of degree $k$.

Finally it is rather well-known that any homogeneous polynomial is the `diagonalization´ of a $k$-linear form. See here or the book on differential calculus by Henri Cartan.

Answer 2

Since $f\in C^k(\mathbb R^m),$ Taylor's theorem in several variables shows that

$$f(x) = P(x) + o(|x|^k)\,\, \text {as } x\to 0,$$

where $P$ is a polynomial of degree no more than $k.$ We can write $P=\sum_{j=0}^{k}P_j,$ where $P_j$ is a homogenous polynomial of degree $j, j=0,\dots, k.$

Lemma: Suppose $C$ and $a_0,a_1,\dots,a_k$ are constants such that

$$Cr^k = \sum_{j=0}^{k}a_jr^j + o(r^k)\,\, \text {as } r\to 0^+.$$

Then $C=a_k.$

I'll leave the proof of the lemma to you for now. Ask if you have questions.

Let $S$ be unit sphere in $\mathbb R^m.$ For each fixed $\zeta \in S,$ we have

$$f(r\zeta) = r^kf(\zeta) = \sum_{j=0}^{k}r^j P_j(\zeta) + o(r^k),\,\,r>0.$$

Apply the lemma to conclude $f(\zeta)=P_k(\zeta).$ This is true for all $\zeta\in S,$ so $f=P_k$ on $S,$ hence everywhere in $\mathbb R^m$ by positive $k$-homogeneity. Thus $f$ is a homogeneous polynomial of degree $k.$

But now observe any such polynomial is a finite linear combination of $k$-homogeneous monomials, so it's enough to show such a monomial is a $k$-linear form. To see the latter, consider a special case to get the basic idea: Suppose we're in $\mathbb R^2$ and our monomial is $ x_1^2x_2^3.$ Define

$$\varphi ((x_1,x_2), (y_1,y_2), (z_1,z_2), (u_1,u_2),(v_1,v_2)) = x_1y_1z_2u_2v_2 .$$

Then $\varphi$ is a $5$-linear form on $(\mathbb R^2)^5$ and, with $x=(x_1,x_2),$ we have

$$\varphi (x,x,x,x,x) = x_1\cdot x_1\cdot x_2\cdot x_2\cdot x_2 = x_1^2x_2^3.$$