Find the adjoint of the right shift operator $T$ in $\ell^1$.

More specifically, $T: X \longrightarrow Y$ defined by

$$Tx= T(x_1, x_2, \dots, x_n, \dots) = (0, x_1, x_2, \dots, x_n\dots) = y$$


I fist showed $T$ was linear, then bounded by noticing sum of the series generated by $(|x_n|)$ is the same as the series generated by $(|y_n|)$. Now it remains for me to find the adjoint operator of $T$.

I'm sort of confused at this step though because I think I'm supposed to be finding the adjoint in the sense of $T^\times: Y' \longrightarrow X'$ defined by (in Kreyszig's functional analysis book)

$$f(x) = \left(T^\times g\right)(x) = g(Tx), \text{where } g \in Y'$$

which I don't believe is the same as the Hilbert adjoint (which my brain keeps circling back to). Do I need to find $f$ explicitly here?

More or less I'm hoping someone can clearly explain what the idea is and maybe give me a step in the right direction. Thank you.

  • 3
    $\begingroup$ The evaluation of a functional can still be seen by taking the same dot product, but with an element of $\ell^{\infty}$. $g(x)=g_0x_0+g_1x_1+...$. So $g(Tx)=g_0\cdot0+g_1\cdot x_0+...=(g_1,g_2,...)\cdot (x_0,x_1,...)$, which is the same as evaluating at $x$ the left shift of $g$. $\endgroup$
    – logarithm
    May 8, 2019 at 14:07
  • $\begingroup$ The right shift operator in $\ell^1$ is an isometry and hence it's adjoint (which happens to be the left shift operator) is a co-isometry. $\endgroup$ Jun 16, 2021 at 16:15

3 Answers 3


Let me use a different notation, which really looks like the one used in the Hilbert space setting, i.e. let $\langle g, y \rangle $ denote $g(y)$ the pairing of $g \in Y^\prime$ and $y\in Y$. The adjunct operator of $T:X\to Y$ is then defined as the unique $T^\times :Y^\prime \to X^\prime$ s.t.:

$$\langle T^\times g, x \rangle = \langle g, Tx \rangle \; .$$

I used the angular brackets $\langle \cdot ,\cdot \rangle$ because of two well-known features of $\ell^p$ spaces, namely:

  1. for each $(x_n) \in \ell^p$ and $(f_n) \in \ell^{p^\prime}$ ($p^\prime$ is the conjugate exponent of $p$) one has $(f_nx_n) \in \ell^1$;
  2. each functional $f \in (\ell^p)^\prime$ can be represented by a unique sequence $f=(f_n)\in \ell^{p^\prime}$ s.t. equality: $$f(x) = \sum_{n=1}^\infty f_nx_n =\langle f , x\rangle$$ holds for all $x=(x_n) \in \ell^p$.

Usually such features are summarized in the frase:

$\ell^{p^\prime}$ is the dual of $\ell^p$ and the pairing coincides with the dot-product.

Therefore your problem recasts in a (probably) more familiar form: for each fixed $g=(g_n) \in \ell^\infty = \ell^{1^\prime}$, find the sequence $T^\times g=:f=(f_n) \in \ell^\infty$ s.t.: $$\langle f, x \rangle = \langle g, Tx \rangle $$ for all $x=(x_n) \in \ell^\infty$.

This is exactly what you do when you have to "find" the adjunct operator in the Hilbert space setting, isn't it? ;-)


Since $\ell^1$ is not a Hilbert space there is no such thing as Hilbert adjoint here anyway. $T^{\times}:Y'\to X'$ is defined by $T^{\times}(f)(x_1,x_2,...)=f(T(x_1,x_2,...))=f(0,x_1,x_2,...)$, that's it.


Presumably, the adjoint is supposed to be from $\ell^\infty$ to $\ell^\infty$, as $\ell^\infty$ is the standard way to represent elements of $(\ell^1)'$. In particular, the map $$\ell^\infty \to (\ell^1)' : (y_n)_{n=1}^\infty \in \ell^\infty \mapsto \left((x_n)_{n=1}^\infty \mapsto \sum_{n=1}^\infty x_n y_n\right) \in (\ell^1)'$$ is an isometric isomorphism between the spaces, so we tend to identify the functionals in $(\ell^1)'$ with sequences in $\ell^\infty$.

So, what does the adjoint $T^*$ do to a sequence $(y_n) \in \ell^\infty$? As a functional in $g \in (\ell^1)'$, $g$ sends $(x_n) \in \ell^1$ to $\sum_n x_n y_n$. So, $$g(T(x_n)) = 0 y_1 + x_1 y_2 + x_2 y_3 + \ldots = \sum_{n=1}^\infty x_n y_{n+1}.$$ In particular, note that this functional is represented by the sequence $(y_{n+1})_{n=1}^\infty$. This is the result after applying $T^*$ to $(y_n)_{n=1}^{\infty} \in \ell^\infty$, in other words, $T^*$ is simply a left shift operation (much like in Hilbert Spaces).

Aside: the adjoint in real Banach Spaces is a generalisation of the adjoint in real Hilbert Spaces. In Banach spaces, we have the pairing $$\langle \cdot, \cdot \rangle : X \times X' \to \Bbb{R}$$ that maps $\langle x, y^* \rangle$ to $f(x)$. When $X$ is a Hilbert Space, then $X'$ can be identified with $X$, where every functional $y^*$ can be expressed uniquely as a vector $y \in X$, where $y^* = ( \cdot, y )$ and $( \cdot, \cdot)$ is the Hilbert space inner product (this is the Riesz Representation Theorem). So, $\langle \cdot, y^* \rangle = ( \cdot, y )$, showing how the two bilinear forms are more or less interchangeable.

The Banach space adjoint can be defined using this pairing, by saying, for all $x \in X$ and $x^* \in X'$, $$\langle Tx, x^* \rangle = \langle x, T^*(x^*)\rangle.$$ Note that this agrees with the usual definition, as $$\langle Tx, x^* \rangle = x^*(T(x)) = (T^*x^*)(x) = \langle x, T^* x^* \rangle.$$ But, since this bilinear form, in the Hilbert space case, might as well be the inner product, it also agrees with the usual definition of Hilbert space adjoints.


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