Let $f:\mathbb{R}^n\to\mathbb{R}$ be a twice continuously differentiable function such that $f(tX)=t^2f(X)$ for all $X\in\mathbb{R}$. Show that $f$ is a quadratic form.(You need some formulas of calculus in several variables to do this.)

Quadratic form definition:

Let $V$ be a finite dimensional space over the field $K$. Let $g=\langle ,\rangle$ be a symmetric bilinear form on $V$. By the quadratic form determined by $g$, we shall mean a function:

$f:V\to K$

such that $f(v)=g(v,v)=\langle v,v\rangle$

I think I need to get a bilinear form and prove $f$ fulfils its properties. I have the following formula $g(x,y)=\frac{1}{2}(f(x+y)-f(x)-f(y))$, and it is clear $g(x,x)=f(x)$.


How am I supposed to prove this theorem?


2 Answers 2



Let's take derivatives with respect to $t$. We denote by $D_i$ the derivative with respect to $x_i$.

On the right-hand side the derivative is $2tf(x)$ since $f(x)$ is just a constant with respect to $t$. On the left-hand side we need to apply the chain rule in several variables. $D_t(f(tx))=\sum_i D_t(tx_i)(D_if)(tx)=\sum_ix_i(D_if)(tx)$.

$\sum_i x_i(D_if)(tx)=2tf(x)$

and again


Now put $t=0$.

This means that $f(x)=\sum_{i,j}C_{i,j}x_ix_j$, where $C_{i,j}=\frac{1}{2}(D_{i,j}f)(0)$.

  • $\begingroup$ I am sorry but could you be more explicit in your terminology please? $\endgroup$ Jul 27, 2017 at 20:51
  • $\begingroup$ @PedroGomes Sure. Which part? $\endgroup$
    – Hellen
    Jul 27, 2017 at 20:51
  • $\begingroup$ I am not understanding what you are doing here $\sum_i x_i(D_if)(tx)=2tf(x)$ $\endgroup$ Jul 27, 2017 at 20:53
  • $\begingroup$ @PedroGomes Added a paragraph. It is just derivative with respect to $t$. You need to use chain rule in several variables to get the left-hand side. $\endgroup$
    – Hellen
    Jul 27, 2017 at 21:02
  • $\begingroup$ Thanks a lot! I got it! $\endgroup$ Jul 27, 2017 at 21:04

Fix $x$ any point in $\mathbb{R}^n$. We have that $g(t)=f(tx)-t^2f(x)$ is constant. Hence, $g''(t)=0$ for all $t$.

Computing, we have that $$0=g''(t)=\langle \mathrm{(Hess}_{tx} f)\cdot x, x\rangle-2f(x)$$ for all $t \in \mathbb{R}$. Taking $t=0$ yields that $$0=\langle \mathrm{(Hess}_{0} f)\cdot x, x\rangle-2f(x),$$ which implies $$f(x)=\frac{1}{2}\langle \mathrm{(Hess}_{0} f)\cdot x, x\rangle.$$ Since $x$ is arbitrary, we have the result.


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