Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace

Question

Let $(X, \mu)$ be your favourite measure space (finite or $\sigma$-finite if you like), let $g \in L^2$ (say, the scalar field of $L^2$ is $\mathbb{R}$, though this probably doesn't matter). Let $\tilde g: X \to \mathbb{R}$ be a measurable function and assume that there exists a norm dense vector subspace $D$ of $L^2$ with the following property:

For every $f \in D$ the function $\tilde g f$ is integrable, and we have $\int \tilde g f \, d\mu = \int g f \, d\mu$.

Question. Does it follow that $\tilde g = g$ almost everywhere?

Remarks.

• Under the given assumptions, the propertiy $\tilde g = g$ almost everyhwere is equivalent to $\tilde g \in L^2$. For if $\tilde g \in L^2$, then the integral equality in the assumption extends to all $f \in L^2$ by density, and for $f := \tilde g - g$ we thus obtain that $\lVert \tilde g - g\rVert^2 = \int (\tilde g - g) f \, d\mu = 0$.

  In other words, the question asks whether it is impossible to represent, on any dense subspace of $L^2$, a continuous linear functional on $L^2$ by a non-$L^2$-function.

• If $D$ is a lattice ideal in $L^2$, meaning that $f_1 \in D$ whenever $\lvert f_1 \rvert \le \lvert f_2 \rvert$ for some $f_2 \in D$, then the answer to the question is yes.

Proof of the claim in the second bullet point. Assume that $D$ is a lattice ideal, and let $f \in D$. There exists a measurable function $s: X \to \mathbb{R}$ of modulus $\lvert s \rvert = 1$ such that $\tilde g f s \ge 0$. Note that this implies $\lvert \tilde g f \rvert = \tilde g f s$. As $\lvert s f \rvert = \lvert f \rvert$ we have $sf \in D$, so it follows that $$ \label{1}\tag{$*$} \int \lvert \tilde g f \rvert \, d\mu = \int \tilde g f s \, d\mu = \int g f s \, d\mu \le \lVert f \rVert \lVert g \rVert. $$ Now, consider a function $0 \le h \in L^2$. By density, there exists a sequence $(f_n)$ in $D$ which converges to $h$. By replacing each $f_n$ with $(h \land f_n) \lor 0$ (where $\land$ denotes the pointwise minimum and $\lor$ denotes the pointwise maximum of functions) we may assume that $0 \le f_n \le h$ for each $n$. Moreover, by then replacing each $f_n$ with the pointwise maximum of the functions $f_1, \dots, f_n$, we may also assume that the sequence $(f_n)$ is increasing. Hence, it follows from the monotone convergence theorem and from \eqref{1} that $$ \int \lvert \tilde g h \rvert \, d\mu = \lim_{n \to \infty} \int \lvert \tilde g \rvert f_n \, d \mu \le \lim_{n \to \infty} \lVert f_n \rVert \lVert g \rVert = \lVert h \rVert \lVert g \rVert. $$ For a general (i.e., not necessarily positive) function $h \in L^2$ we can apply the estimate that we just proved to $\lvert h \rvert$ and thus obtain the same estimate $$ \int \lvert \tilde g h \rvert \, d\mu \le \lVert h \rVert \lVert g \rVert $$ for even all $h \in L^2$. But this shows that $\tilde g\in L^2$, so $\tilde g = g$ almost everyhwere due to the remark in the first bullet point. $\square$

Answer 1

The answer is no and the following result provides a quite interesting counterexample. This is a known result, but I am not sure where to find it in the literature.

Theorem. If $f\in L^1_{\rm loc}(\mathbb{R}^n)\setminus L^2(\mathbb{R}^n)$, then there is an orthonormal basis $\{\varphi_k\}_{k=1}^\infty$ of $L^2(\mathbb{R}^n)$ consisting of compactly supported smooth functions, such that $$ \int_{\mathbb{R}^n}f(x)\varphi_n(x)\, dx=0 \quad \text{for all $n=1,2,3\ldots$} \tag1 $$

Indeed, if $D=\operatorname{span}\{\varphi_k\}$, then $$ \int_{\mathbb{R}^n} f\varphi = \int_{\mathbb{R}^n} 0\cdot\varphi, \quad \forall \varphi\in D. $$

Proof. Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ be the given field. In the proof we will need the following result.

Lemma. Let $L:X\to\mathbb{K}$ be a discontinuous linear functional defined on a normed space $X$. Then $\ker L$ is a dense linear subspace of $X$.

Proof. Since $L$ is discontinuous, it is unbounded, so there is a sequence $y_n\in X$ such that $|Ly_n|>n\Vert y_n\Vert$. Thus $x_n=y_n/\Vert y_n\Vert$ satisfies $$ \Vert x_n\Vert =1 \quad \text{and} \quad |Lx_n|>n. \tag2 $$ Let $y\in X$ be arbitrary. Then (2) yields $$ \ker L\ni y-\Big(\frac{Ly}{Lx_n}\Big)x_n\to y $$ so $\ker L$ is dense in $X$. $\Box$

Proof of the theorem. $X:=C_c^\infty(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)$ is a dense linear subspace. Note that $$ L:X\to\mathbb{K}, \qquad L\varphi=\int_{\mathbb{R}^n}f(x)\varphi(x)\, dx $$ is a linear functional on $X$. We will show that $L$ is unbounded (with respect to the $L^2$ norm in $X$). Suppose to the contrary that $L$ is bounded. Then it extends uniquely to a bounded linear functional $\tilde{L}:L^2(\mathbb{R}^n)\to\mathbb{K}$. Therefore, the Riesz Representation yields existence of $g\in L^2(\mathbb{R}^n)$, such that $$ \tilde{L}\varphi=\int_{\mathbb{R}^n}\varphi(x)\overline{g(x)}\, dx \quad \varphi\in L^2(\mathbb{R}^n). $$ In particular $$ \int_{\mathbb{R}^n}(f(x)-\overline{g(x)})\varphi(x)\, dx=0 \quad \text{for all $\varphi\in C_c^{\infty}(\mathbb{R}^n)$.} $$ Therefore $f-\overline{g}=0$ a.e., $f=\overline{g}\in L^2(\mathbb{R}^n)$ which contradicts the assumption about $f$.

We proved that $L$ is discontinuous, hence by the lemma, $\ker L\subset X=C_c^\infty(\mathbb{R}^n)$ is dense and hence it is dense in $L^2(\mathbb{R}^n)$. Now applying the Gramm-Schmidt orthogonalization to a countable and dense subset of $\ker L$ we obtain an orthonormal basis $\{\varphi_k\}_{k=1}^\infty\subset\ker L$ of $L^2(\mathbb{R}^n)$ consisting of compactly supported smooth functions. Since $L\varphi_k=0$ for all $k$, (1) follows. $\Box$

Answer 2

I believe the following is a simple counterexample:

Let $(X,\mu)$ be $\mathbb{N}$ with the counting measure (so I will be writing $\ell^2$ for $L^2(X,\mu)$. Let $g=0$ and $\tilde g(n) = (-1)^n$. Let $D$ be the set of (finite) linear combinations of sequences of the form $(0,\ldots,0,1,1,0,\ldots)$ (i.e., two consecutive $1$'s). Clearly, if $f\in D$ then $\sum_n f(n)\,g(n)$ is convergent, in fact, it is even a finite sum, and its value is $0$ because this holds on the elements spanning $D$.

Now $D$ is dense in $\ell^2$ (for the norm topology): to see this, it is enough to show $D^\perp = 0$ (orthogonal in $\ell^2$). But if $h \in D^\perp$ then we have $h(n) = -h(n+1)$ for every $n$, so $h$ is proportional to $\tilde g$, which is not in $\ell^2$, so $h=0$.