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Okay: $\|Tf\|_2^2=\int_a^b|\sum_{j=1}^n\int_a^b\phi_j(t)\psi_j(s)f(s)ds|^2dt\le\int_a^b|\sum_{j=1}^n|\phi_j(t)|\int_a^b|\psi_j(s)f(s)|ds|^2dt$ $\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\int_a^b|\sum_{j=1}^n \phi_j(t)|^2dt\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\|\sum_{j=1}^n \phi_j(t)\|_2^2$ Thus: $\|Tf\|\le\sup_j\|\psi_j\|_2\|\sum_{j=1}^n \phi_j(t)\|_2\|f\|_2$ I.e. $T$ ...


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Let $H$ be a complex Hilbert space and let $T \in B(H)$ be an isometry. We claim that $T^\ast T = I$. To this end let $h \in H$ and note that by the Hilbert space Riesz representation theorem a linear functional in $H^\ast$ corresponds to an element $h \in H$ ($h \mapsto \langle \cdot, h \rangle$). Also note that if $\varphi (x) = 0$ for all $\varphi \in ...


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1) For every $n$, $x_n=(\underbrace{1,\ldots,1}_{n\mathrm{\ times}},0,\ldots)\in c_{00}$, $\Vert x_n\Vert=1$ but $x_n$ maps to $n$, so this function is not continuous 2)Let $\phi:\ell^\infty \ni (x_n)\mapsto (\frac{1}{n+1}x_n)\in \ell^2$. Since $$\Vert ...


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Here is a sketch of the first part. For each $f\in C[0,1]$, \begin{eqnarray*} \|Tf\| &=& \max_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)f(s)ds\right|\\ &\leq& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)f(s)\right|ds\\ &\leq& \|f\|_{\infty}\max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\ \end{eqnarray*} Therefore $\|T\|$ is ...



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