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I have an operator $A: \ell_1 \to \ell_1, Ax = (x_1+x_2, x_1-x_2, x_3,...,x_k,...)$

AFAIK, norm of $\ell_1$ is $\sum_{n=1}^{\infty}|x_n|$

How to find a norm of this operator?

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Norm should be 3. – Mhenni Benghorbal May 13 '14 at 7:58
@MhenniBenghorbal why? How to prove it? – michaeluskov May 13 '14 at 7:59
up vote 3 down vote accepted


\begin{align*} \newcommand{\norm}[1]{\left\| #1 \right\|_1} \newcommand{\abs}[1]{\left| #1 \right|} \left\| A \right\| &= \sup_{x \in \ell^1} \frac{\norm{Ax}}{\norm{x}} \\ &= \sup_{x \in \ell^1} \frac{\abs{x_1 + x_2} + \abs{x_1 - x_2} + \displaystyle \sum_{i=3}^ \infty |x_i|} {\abs{x_1} + \abs{x_2} + \displaystyle \sum_{i=3}^\infty |x_i|} \\ &= \sup_{x, y, z \in \mathbb{R}} \frac{\abs{x + y} + \abs{x - y} + \abs{z}} {\abs{x} + \abs{y} + \abs{z}} \end{align*}

Further Hint

Triangle inequality: $|x + y| + |x - y| \le 2|x| + 2|y|$

Now try $x, y, z = 1, 0, 0$.

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But is there supremum of this function? I think no, WolframAlpha says no either, but how to prove it? – michaeluskov May 13 '14 at 7:51
@user23791 I added a further hint. – 6005 May 13 '14 at 7:56
Sorry, I didn't understand your further hint. So this function will be equal to 2 in $(1,0,0)$. And what does it mean? – michaeluskov May 13 '14 at 8:47
@user23791 The supremum exists and is achieved. To show this, you need to prove (1) the expression is bounded above by the maximum (see "Triangle inequality" of my further hint) and (2) the maximum is attained. – 6005 May 13 '14 at 8:56
@user23791 yep. – 6005 May 13 '14 at 9:34

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