Let \(f:\mathbb R^n\to\mathbb R\) satisfy:
The function \[ g(x,y)=f(x+y)-f(x)-f(y) \] is bilinear.
For every \(x\in\mathbb R^n\) and \(t\in\mathbb R\), \[ f(tx)=t^2f(x). \]
Show that there is a linear transformation \(A:\mathbb R^n\to\mathbb R^n\) such that \[ f(x)=\langle x,Ax\rangle \] for the usual inner product.
Solution
Let \[ e_1,\ldots,e_n \] be the standard basis of \(\RR^n\).
1. One has\[ f(0)=0. \]
Apply the homogeneity hypothesis with \(t=0\): \[ f(0) = f(0x) = 0^2f(x) =0. \]
2. For every \(x\in\RR^n\),\[ g(x,x)=2f(x). \]
By definition, \[ g(x,x) = f(2x)-2f(x). \] The homogeneity hypothesis with \(t=2\) gives \[ f(2x)=4f(x). \] Therefore \[ g(x,x) = 4f(x)-2f(x) = 2f(x). \]
3. Define the real matrix\[ A=(a_{ij}) \]by\[ a_{ij} \coloneqq \frac12g(e_i,e_j). \]Then \(A\) defines a linear transformation\[ A:\RR^n\longrightarrow\RR^n. \]
Every real \(n\times n\) matrix defines a linear endomorphism of \(\RR^n\) in the standard basis.
4. For every \(x\in\RR^n\),\[ \langle x,Ax\rangle = \frac12g(x,x). \]
Write \[ x=\sum_{i=1}^n x_i e_i. \] Then \[ \begin{aligned} \langle x,Ax\rangle &= \sum_{i,j=1}^n x_i a_{ij}x_j\\ &= \frac12 \sum_{i,j=1}^n x_ix_jg(e_i,e_j). \end{aligned} \] Since \(g\) is bilinear, \[ \sum_{i,j=1}^n x_ix_jg(e_i,e_j) = g(x,x). \] This gives the identity.
5. For every \(x\in\RR^n\),\[ \boxed{ f(x)=\langle x,Ax\rangle }. \]
Step 2 gives \[ f(x)=\frac12g(x,x), \] and step 4 identifies the right side with \(\langle x,Ax\rangle\).
6. Q.E.D.
The linear transformation constructed in step 3 has the required property by step 5.