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Let $f \in L^1[0,1]$, but $f \notin L^2[0,1]$. Consider the subspace $X$ of $L^2[0,1]$ such that $X= \{\phi \in L^2[0,1]: \int f \phi = 0\}$. Want to show that $X$ is dense in $L^2[0.1]$. I tried proving this by checking that $\langle f, g \rangle, \forall g \in X$ implies $f = 0$. However, this does not seem like a plausible way of doing this. I am also hinted by using the theory of densely defined operator, but I only know of this by its definition.

mechanodroid
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penny
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1 Answers1

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Let $D\subset L^2[0,1]$ be the set of $\phi\in L^2[0,1]$ such that $f\phi$ is integrable. Note that $D$ is dense in $L^2[0,1]$ (since $f\in L^1[0,1]$, $D$ contains all of $L^\infty[0,1]$ which is dense in $L^2[0,1]$).

We can now consider the functional $T$ on $D$ given by $T\phi=\int f\phi$. The subspace $X$ is the kernel of $D$. Since $f\not\in L^2[0,1]$, $T$ must be unbounded. But the kernel of any unbounded linear functional on a normed vector space is dense. Thus $X$ is dense in $D$, and hence also in $L^2[0,1]$.

Eric Wofsey
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  • Why is $T$ unbounded? Is it by Holder? – penny Sep 14 '18 at 21:57
  • Uh, maybe, depending on exactly what your statement of Holder is. If $T$ were bounded, it would extend to a bounded functional on $L^2$, which must be given by integrating against some function $g\in L^2$. But then $g$ and $f$ must be the same, since their integrals against any bounded function are the same, so $f\in L^2$. – Eric Wofsey Sep 14 '18 at 22:15