# Surjectivity of linear functonal from $L_2[-1,1]$

In studying for an exam I came across a problem that claimed that given a non-zero continuous linear functional $$F:L_2[-1,1] \to \mathbb{R}$$ we immediately know that $$F$$ is surjective by Riesz representation theorem. The representation theorem implies there exists a unique non-zero $$f$$ such that $$F(g) = \langle g,f\rangle$$ for all $$g\in L_2[-1,1]$$. How is it clear that $$F(g)$$ is surjective? Given a real number $$c$$, I need to find a function g such that $$\int_{-1}^1 g f = c$$ Any hints would be much appreciated.

Thanks.

• Right that's why I said that it won't work – 1729 May 4 '20 at 14:26

Let $$||\cdot||$$ be the $$L^2[-1,1]$$-norm. The Riesz representation theorem gives a non-zero $$f$$ such that $$F(g)=\langle g,f\rangle=\int_{[-1,1]}|fg|$$ for some $$0\neq f\in L^2[-1,1]$$.
Assume that $$||f||=0$$, then $$f=0$$ almost everywhere on $$[-1,1]$$. Then, in particular $$F(g)=0$$ for all $$g\in L^2[-1,1]$$. However we assumed that $$F$$ was a non-zero linear functional. Hence $$||f||\neq 0$$.
Now, we have $$F(f)=\langle f,f\rangle=||f||^2$$, hence $$F(af/||f||^2)=a\in \mathbf{R}$$. We conclude that $$F$$ is surjective.
Let $$g = \frac{cf}{||f||_2^2}$$
Now $$||f||_2^2\neq 0$$ since other wise $$F(g)= \langle g,f\rangle$$ would imply $$F$$ is zero functional ($$||f||_2^2 = 0 \Rightarrow f=0$$ almost everyhwere so $$\langle g,f\rangle=0$$).
If $$V$$ is a real vector space and $$T:V\to \mathbb R$$ is a nonzero linear functional, then $$T$$ is surjective. This has little to do with continuity, $$L^2,$$ Riesz Representation. It follows from the simple fact that if $$x\in \mathbb R,x\ne 0,$$ then $$\{\lambda x: \lambda \in \mathbb R\}= \mathbb R.$$