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Find the adjoint operator of $T(x_1,x_2,x_3,\dots)=(\sum_{n=2}^\infty x_n, x_1, x_2, x_3, \dots)$ in $\ell^1$.

I know that the dual space of $\ell^1$ is $\ell^\infty$ and thus $T^*$ should map from $\ell^\infty$ to $\ell^\infty$.

Then I try to see what $T^*$ does to any $a \in \ell^\infty$. $T^*a=aT$ by the definition of adjoint operators in Banach spaces.

However, what I want to know is the element to which $T^*$ maps $a$. I don't know how to play with it to find the adjoint.

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    $\begingroup$ Try to write out the infinite matrix representation of $T$ in the standard basis. What is the transpose of this matrix? $\endgroup$ Commented Mar 2, 2017 at 18:43

1 Answer 1

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Let $x\in l^1, y \in l^\infty$, then

$$\langle Tx, y \rangle = \langle (\sum_{n=2} x_n, x_1, x_2, \cdots ), (y_1,y_2,\cdots) \rangle $$

$$ = y_1 \sum_{n=2} x_n + x_1 y_2 + \cdots$$

$$ = x_1 y_2 + x_2(y_1+y_3) + x_3 (y_1 + y_4) + \cdots$$

$$ = \langle (x_1 , x_2, \cdots), (y_2, y_1+y_3 , y_1+y_4 , +\cdots )\rangle$$

Hence

$$T^*(y_1, y_2, y_3, \cdots) = (y_2, y_1+y_3, y_1+y_4, \dots)$$

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