Let \(F\) be a field. We have considered certain special linear functionals on \(F[x]\) obtained via ‘evaluation at \(t\)’: \[ L(f) = f(t). \] Such functionals are not only linear but also have the property that \(L(fg) = L(f)L(g)\). Prove that if \(L\) is any linear functional on \(F[x]\) such that \[ L(fg) = L(f)L(g) \] for all \(f\) and \(g\), then either \(L = 0\) or there is a \(t\) in \(F\) such that \(L(f) = f(t)\) for all \(f\).
Solution
Theorem.
If \(L:F[x]\to F\) is linear and
multiplicative, then either \(L=0\) or
\(L(f)=f(a)\) for a unique \(a\in F\).
- If \(L=0\), then \(L\) is the zero functional, which is the first alternative in the statement of the theorem, so the claim holds.
- Assume \(L\neq0\). Since \(L\) is linear and \(L(1)=L(1\cdot1)=L(1)^2\), we get \(L(1)=1\).
- Let \(a=L(x)\in F\). Then \[ L((x-a)g)=L(xg)-aL(g)=L(x)L(g)-aL(g)=0 \] for every polynomial \(g\).
- Define \(\varphi:F[x]\to F\) by \(\varphi(f)=f(a)\). For monomials, \((x-a)g(a)=0\), so \(\varphi((x-a)g)=0\) for all \(g\).
- Every \(f\in F[x]\) has unique decomposition \(f(x)=q(x)(x-a)+c\) with \(c=f(a)\). Applying \(L\) gives \(L(f)=L(q(x)(x-a))+L(c)=c=L(a)\).
- Therefore \(L(f)=f(a)\) for all polynomials \(f\).
Hence the claim follows. ∎