Tag Info

Hot answers tagged

2

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 ...


2

$Tf=\phi*f$, 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 ...


1

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible