# Identity operator on $L^2(\mathbb{R}^d)$

I want to show that the identity operator on $L^2(\mathbb{R}^d)$ cannot be given by an absolutely convergent integral operator. That is, if $K(x,y)$ is a measurable function on $\mathbb{R}^d \times \mathbb{R}^d$ such that for each $f \in L^2(\mathbb{R}^d)$ and $T(f)(x) = \int_{\mathbb{R}^d} K(x,y)f(y)dy$ converges for almost every $x$, then $T(f) \neq f$ for some $f$.

Therefore, suppose that $T(f)(x)$ converges absolutely for almost every $x$ and $T(f) = f$ for all $f$. Then $$f(x) = \int_{\mathbb{R}^d} K(x,y) f(y) dy \leq \int_{\mathbb{R}^d} \left| K(x,y)f(y) \right|dy< \infty.$$ I don't really know how to proceed from here.

• I think the point seems to be that if $\int K(x,y)^2 dxdy<\infty$ then $T$ is necessarily Hilbert-Schmidt, whereas if $\int K(x,y)^2 dxdy=+\infty$ then the operator is necessarily unbounded (but I can't seem to find an easy proof). The identity operator on $L^2$ clearly falls into neither category. – Shalop Mar 23 '17 at 6:12

## 1 Answer

Here is an argument based on the heuristic observation that the essential support of $K$ must be contained in the diagonal, which is itself a measure zero set.

Part 1, The Approximation: Let $f(x)=(2\pi)^{-d/2}e^{-\lvert x \rvert^2/2}$ and define $$\tilde{f}(x) = \begin{cases} \int \lvert K(x,y)f(y)\rvert\,dy&\mbox{ if } \int \lvert K(x,y)f(y)\rvert\,dy<\infty,\\ 0&\mbox{ otherwise }. \end{cases}$$ Then $\tilde{f}$ is measurable. Consider the measurable set $E_t=(\tilde{f}\leqslant t)$ and define $\phi_t=1_{E_t}\cdot f$ and $K_t(x,y)=\phi_t(x)K(x,y)\phi_t(y)$. Note that $\phi_t\rightarrow f$ pointwise almost everywhere as $t\rightarrow\infty$. We have $$\int \lvert K_t(x,y)\rvert \, dx dy \leqslant \int t\cdot \phi_t(x) \, dx = t <\infty,$$ i.e. $K_t$ is integrable.

Interlude, The Diagonal: Let $\Delta=\{(x,x)\in\mathbb{R}^{2d}\;\vert\; x\in\mathbb{R}^d\}$ denote the diagonal. Pick open sets $U,V\subseteq\mathbb{R}^{d}$ such that $U\cap V=\emptyset$, let $\Sigma$ denote the Borel $\sigma$-algebra on $\mathbb{R}^{2d}$, let $\Sigma_{U\times V}=\{(U\times V)\cap A\;\vert\; A\in\Sigma\}$ denote the trace $\sigma$-algebra on $U\times V$.

Part 2, The Dynkin Class Argument: Define the family $$\mathcal{D}=\left\{ A\in \Sigma_{U\times V} \; \middle| \; \int_A 1_{U\times V}(x,y)K_t(x,y)\,dxdy=0 \right\}.$$ We will now argue that $\mathcal{D}$ is a Dynkin system. We have

1. $\int_{U\times V} 1_{U\times V}(x,y)K_t(x,y)\,dxdy =\langle 1_U\cdot \phi_t,1_V\cdot \phi_t\rangle = 0$.
2. If $A\in\mathcal{D}$, then $\int_{(U\times V)\setminus A} 1_{U\times V}(x,y)K_t(x,y)\,dxdy = -\int_{A} 1_{U\times V}(x,y)K_t(x,y)\,dxdy = 0$.
3. If $(A_j)_{j\in\mathbb{N}}$ is a sequence of pairwise disjoint elements of $\mathcal{D}$, then $$\int_{\bigcup_{j=1}^\infty A_j} 1_{U\times V}(x,y)K_t(x,y)\,dxdy = \sum_{j=1}^\infty\int_{A_j} 1_{U\times V}(x,y)K_t(x,y)\,dxdy = 0.$$

Next, we define the family $$\mathcal{P}=\left\{ A\in\Sigma_{U\times V} \; \middle| \; A=B\times C \right\},$$ which is clearly a $\pi$-system. Furthermore, if $B\times C\in\mathcal{P}$, then $$\int_{B\times C} 1_{U\times V}(x,y)K_t(x,y)\,dxdy =\langle 1_{B}\cdot \phi_t,1_{C}\cdot \phi_t\rangle = 0,$$ so $\mathcal{P}\subseteq\mathcal{D}$. It follows from Dynkin's Theorem that $\sigma(\mathcal{P})\subseteq\mathcal{D}$, where $\sigma(\mathcal{P})$ denotes the $\sigma$-algebra generated by $\mathcal{P}$. But the $\sigma$-algebra generated by $\mathcal{P}$ is the Borel $\sigma$-algebra on $U\times V$; it follows that $K_t=0$ almost everywhere in $U\times V$.

Conclusion: Since we may cover $\Delta^c$ by a countable family of open sets of the form $U\times V$ where $U,V\subseteq\mathbb{R}^d$ are open and satisfy $U\cap V=\emptyset$, we conclude that $K_t=0$ almost everywhere in $\Delta^c$, and therefore $K_t=0$ almost everywhere in $\mathbb{R}^{2d}$. Since $K_n(x,y)\rightarrow f(x)K(x,y)f(y)$ for almost every $(x,y)\in\mathbb{R}^{2d}$ as $n\rightarrow \infty$, we conclude that $K=0$ almost everywhere in $\mathbb{R}^{2d}$. Contradiction!