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

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

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

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

## Attempt

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.

• 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 $g$. May 8, 2019 at 14:07
• 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. Jun 16, 2021 at 16:15

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.