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Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy$$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz.$$ This is just the convolution operator $$f\mapsto \phi*f$$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$ Now one of the fundamental inequalities (not hard to prove) about ...

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$Tf=\phi*f$, where $\phi(t)=e^{-t}\chi_{(0,\infty)}(t)$. (I had $\phi$ backwards in the first version; noticed that guest's $\phi$ was different, then noticed his was right.) So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In ...

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First show that $C^{\frac{1}{2}}(.,V)$ is a norm: $C^{\frac{1}{2}}(\lambda u)=|\lambda|C(u,V)$ and $\{\sum_k ||P_{W_k}(u+v)||^2\}^{\frac{1}{2}}\leq\{\sum_k ||P_{W_k}(u)||^2\}^{\frac{1}{2}}+\{\sum_k ||P_{W_k}(v)||^2\}^{\frac{1}{2}}$ if and only if $\sum_k Re <P_{W_k}u,P_{W_k}v>\leq\{\sum_k ||P_{W_k}(u)||^2\}^{\frac{1}{2}}\{\sum_k ... 1 First check, that$Q^{\frac{1}{2}}$is a norm. The triangle inequality needs some calculation (...$Q(x+y)=\sum_{k=1}^K |<x,g_k>|^2+2Re<x,g_k>\overline {<y,g_k>} +|<y,g_k>|^2$and$(\sum_{k=1}^K Re<x,g_k>\overline {<y,g_k>})^2\leq\sum_{k=1}^K|<x,g_k>|^2\sum_{k=1}^K|<y,g_k>|^2\$, which follows from Cauchy ...

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