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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 at 7:58
    
@MhenniBenghorbal why? How to prove it? –  michaeluskov May 13 at 7:59

1 Answer 1

up vote 3 down vote accepted

Hint

\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 at 7:51
    
@user23791 I added a further hint. –  Goos May 13 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 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. –  Goos May 13 at 8:56
1  
@user23791 yep. –  Goos May 13 at 9:34

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