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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.

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

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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 $\langle g,f\rangle=0$).

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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 $\{\lambda x: \lambda \in \mathbb R\}= \mathbb R.$

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