$$X=\ell_1, \ Y=\ell_2, \ A(x_1,x_2,\ldots )=(y_1,y_2,\ldots)$$ Operator is defined as $y_1=x_1,y_n=x_n-x_{n-1}, \ n=2,3,\ldots $

Prove that $A\in L(X,Y)$ and calculate $\|A\|$.

First, I check if the operator is linear \begin{align*} A((x_1,x_2,\ldots )+\lambda (y_1,y_2,\ldots ))&=A(x_1+\lambda x_2,x_2+\lambda y_2,\ldots )\\ &=(x_1+\lambda y_1, x_2-x_1+\lambda y_2-\lambda y_1, \ldots )\\ &=(x_1+\lambda y_1,(x_2-x_1)+\lambda(y_2-y_1),\ldots )\\ &=A(x_1,x_2,\ldots) +\lambda A(y_1,y_2,\ldots ) \end{align*} next, I check if the operator is bounded (bounded linear operator is continuous) \begin{align*} \| A(x_1,\ldots )\|&=\|(x_1,x_2-x_1,\ldots )\|\\ &= |x_1|^2+\sum_{k=2}^\infty |x_k-x_{k-1}|^2\\ \end{align*} I could use some help on the boundedness and finding $\|A\|$.

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    $\begingroup$ Why did you delete the previous incarnation, and then reask the exact same question? This is not best behavior, you know! Please explain yourself. $\endgroup$ – Jyrki Lahtonen May 8 '15 at 11:51

If $(x_1,x_2,\ldots,x_n)$ is any vector of reals, then $$ \begin{aligned} x_1^2+(x_1-x_2)^2+\cdots+(x_{n-1}-x_n)^2&=2\sum_{i=1}^nx_i^2-2\sum_{i=1}^{n-1}x_ix_{i+1}-x_n^2\\ &\le2\sum_{i=1}^nx_i^2+4\sum_{1\le i<j\le n}|x_i||x_j|\\ &=2\left(\sum_{i=1}^n|x_i|\right)^2. \end{aligned} $$ This implies that for all sequences $x=(x_n)\in\ell^1$ we have $$ ||Ax||_2\le\sqrt2||x||_1. $$ In particular $Ax\in\ell^2$. The boundedness and continuity of $A$ follow from this as well as the inequality $||A||\le\sqrt2$.

But with a unit pulse $x=(1,0,0,0,\ldots)$ we get $Ax=(1,-1,0,0,\ldots)$. Here $||x||_1=1$ and $||Ax||_2=\sqrt2$. Therefore $||A||=\sqrt2$.

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