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Let the mapping $T:\ell^{2}\rightarrow \ell^{2}$ is defined as follow. $$T(x_1,x_2,\ldots,x_n,\ldots)=(x_1,\dfrac{1}{2}x_2,\ldots,\dfrac{1}{n}x_n,\ldots)$$ In this case, i've easily earned: $$\sigma(T)=\{0\}\cup\{\dfrac{1}{n}:n\in\mathbb{N}\}$$ Now, if $\lambda=x+iy\in\mathbb{C}$ and $\lambda\notin\sigma(T)$, i should prove $$\Vert(\lambda I-T)^{-1}\Vert=\dfrac{1}{\displaystyle\inf_{n\in\mathbb{N}}\,\left\vert\lambda-\dfrac{1}{n}\right\vert}$$ can you help me for starting of prove? Thanks.

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You have $$ (\lambda I-T)(x_1,x_2,\ldots)=((\lambda-1)x_1, (\lambda-\frac12)x_2,\ldots). $$ Define an operator $S $ by $$S (x_1,x_2,\ldots)=(\frac{x_1} {\lambda -1},\frac {x_2}{\lambda -\frac12},\ldots). $$ By the choice of $\lambda $ the linear operator $S $ is well-defined and bounded: by construction, $S (\lambda I-T)=(\lambda I-T)S=I $. So $S=(\lambda I-T)^{-1} $ (bounded, because $\lambda\not\in\sigma (T) $. And $$\tag{1}\|S\|=\sup\left\{\frac1 {\left|\lambda-\frac1n\right|}:\ n\in\mathbb N\right\}=\frac1 {\inf\left\{\left|\lambda-\frac1n\right |:\ n\in\mathbb N\right\}}. $$

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The only piece left hanging, to justify the first equality in $(1)$, is to show that if $$ R (x_1,x_2,\ldots)=(r_1x_1,r_2x_2,\ldots) $$ then $\|R\|=\sup\{|r_n|:\ n\} $. This follows, if we write $c=\sup\{|r_n|:\ n\} $: $$ \|Rx\|^2=\sum_j|r_jx_j|^2\leq c^2\sum_j|x_j|^2=c^2\,\|x\|^2, $$ so $\|R\|\leq c$. Let $j$ such that $|r_j|>r-\frac1j$. If $e_j\in\ell^2$ is the sequence with a $1$ in the $j^{\rm th}$ position and zeroes elsewhere, then $\|e_j\|=1$ and $$ \|Re_j\|=|r_j|>r-\frac1j. $$ So $\|R\|=c=\sup\{|r_n|:\ n\}$.

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