# Norm of a singular integral operator

Let $H$ be the characteristic function of $(0,+\infty)$ and let us define for $(x,y)\in \mathbb R^2$, $x\not=y$ $$k(x,y)=\frac{H(x+y)}{iπ(x-y)}.$$ For $u\in C^1_c(\mathbb R)$, we define for $x\in \mathbb R$, $(Ku)(x)=\lim_{\epsilon\rightarrow 0}\int_{\vert x-y\vert>\epsilon} k(x,y) u(y) dy.$ It is not difficult to prove that the operator $K$ can be extended to a bounded operator on $L^2(\mathbb R)$: we have $$k(x,y)=\underbrace{H(x) k(x,y)H(y)}_{H(x)\frac{1}{iπ(x-y)}H(y)} +\underbrace{H(-x) k(x,y)H(y)}_{-\frac{H(x+y)H(-x) H(y)}{iπ\vert x+y\vert}} +H(x) k(x,y)H(-y),$$ and the first term gives rise to a bounded operator (operator-norm 1) since $K_1$ below is bounded, the second (and third) term to a a bounded operator (operator norm$\le 1$) since the operator with non-negative kernel $H(x)H(y)/π(x+y)$ has norm 1, and this gives also the boundedness of the last term. Working a bit around these identities gives that the selfadjoint operator $K$ is smaller than $(1+\sqrt 2)/2$, unfortunately larger than 1.

My question: I believe that the operator-norm of $K$ is 1 (as an operator from $L^2(\mathbb R)$ to itself), but I am not able to prove it, not even able to prove that $K\le 1$. Note that the operator $K_1$ with singular kernel $1/{iπ(y-x)}$ is the standard Hilbert transform, the Fourier multiplier by $\text{sign}(\xi)$, i.e. $$\widehat{K_1 u}(\xi)=\text{sign}(\xi)\hat u(\xi).$$