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Let $f,g\in L^1(\Omega), \Omega\in\mathbb{R}^n$. Define $M=\{ \phi\in C_c^{\infty}(\Omega): \langle f, \phi\rangle_{L^2}=0\}$.

I want to prove the following statement: $\langle g,\phi\rangle_{L^2}=0$ for all $\phi\in M$ implies that there exists $\lambda\in\mathbb{R}$ so that $$g(x)=\lambda f(x)$$ almost everywhere in $\Omega$.

I'm missing a proof idea. Is it worthwhile to consider $h=g-\lambda f$? Any help is greatly appreciated!

DominikS
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john_psl1298
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2 Answers2

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We make use of the fact that $L^2(\Omega)$ is a Hilbert space, and $C_c^\infty(\Omega)\subset L^2(\Omega)$ is dense.

$M$ is a dense subset of $\operatorname{span}\{f\}^\perp$, see the proof below. Therefore $\operatorname{cl}M = \operatorname{span}\{f\}^\perp$ (with $\operatorname{cl}$ being the closure).

There holds $M^\perp = (\operatorname{cl}M)^\perp$. Therefore $M^\perp = \operatorname{span} \{f\}^{\perp\perp} = \operatorname{span}\{f\}$.

Appendix: Density of $M$ in the orthogonal complement. There might be more elegant proofs, but here's one possible approach: Let $g\in\operatorname{span} \{f\}^{\perp}$ (and not in $C_c^\infty$). Then there are two sequences in $C_c^\infty$:

  • $\phi_n \to g$ in $L^2$, and
  • $\psi_n \to f$ in $L^2$.

Both follows from the already stated fact that $C_c^\infty(\Omega)\subset L^2(\Omega)$ is dense. The $\phi_n$ can be decomposed orthogonally into $$\phi_n = \mu_n + \lambda_n f,\quad \text{ for }\mu_n\in f^\perp\text{ and }\lambda_n\in\mathbb C,$$ and $\lambda_n\to 0$, because $\phi_n\to g \in f^\perp$. The only remaining problem is that the $\mu_n\notin C_c^\infty(\Omega)$ necessarily. This is where we use $\psi_n$ to further decompose the $\phi_n$: First note that the sequence $(\psi'_n)$ also satisfies $\psi_n'\to f$: $$\psi'_n:=\frac{\|f\|^2}{\langle \psi_n, f\rangle}\psi_n.$$ We use this in the decomposition of $\phi_n$: $$\phi_n = [\mu_n + \lambda_n(f-\psi_n')] + \lambda_n\psi_n'.$$

  • The term in the square brackets is in $C_c^\infty$ because $\phi_n$ and $\lambda_n\psi_n'$ are.
  • The term in the square brackets is orthogonal to $f$ (because we have normalized $\psi'$ accordingly).
  • The term $\lambda_n\psi_n'\to 0$.

Therefore, also $\phi_n':= \mu_n + \lambda_n(f-\psi_n')\in C_c^\infty \cap \{f\}^\perp$ converges to $g$. Hence $M\subset\{f\}^\perp$ is dense.

DominikS
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  • Thank you for your answer! I just do not underastand why $M$ has to be a dense subset of $span{f}^\perp$. – john_psl1298 Oct 25 '23 at 15:59
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    I have edited the post. Your scepticism is indeed justified: There are examples of Hilbert spaces $H$ where $M\subset H$ dense, and $L\subset H$ a closed linear subspace, but $M\cap L = {0}$. See https://math.stackexchange.com/questions/2128275/is-intersection-of-a-dense-subspace-and-a-closed-subspace-dense-in-the-subspace for example. – DominikS Oct 25 '23 at 16:56
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This is essentially a statement in linear algebra. Indeed, the algebraic mechanism behind all this is the following: if $L_1$ and $L_1$ are linear functionals on a vector space $V$, and $\ker L_1\subset \ker L_2$, then $L_2=\lambda L_1$ for some scalar $\lambda$. (See for example here).

In the case at hand, the vector space is $C^\infty_c(\Omega)$ and $$ \begin{array}{cc} L_1\phi=\langle f, \phi\rangle, & L_2\phi=\langle g, \phi\rangle. \end{array} $$ By definition $M=\ker L_1$ and by assumption $M\subset \ker L_2$. The algebraic lemma mentioned above yields $L_2=\lambda L_1$. Now the density of $C^\infty_c(\Omega)$ into $L^1(\Omega)$ implies that $g=\lambda f$ almost everywhere.

P.S.: This reminded me a recent conversation I had with Alex Iosevich. He was patiently explaining me a certain result, and I unwisely remarked "that's just linear algebra". He immediately replied that my comment applies to virtually ALL mathematics. :-)

Giuseppe Negro
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