# Adjoint operator $T^{\ast}$ in space $l^1$

Find the adjoint operator of $$T:l^1 \rightarrow l^1$$, $$(x_k)_{k\in \mathbb{N}}\mapsto (\sum^{\infty}_{k=1}x_k,0,0,...)$$

In our lecture we defined the adjoint operator as \begin{align*} T^{\ast}(y^{\ast}):X\rightarrow \mathbb{K}\\ x\mapsto y^{\ast}(Tx) \end{align*}

so I think I must do something like this...

\begin{align*} (T^{\ast}y^{\ast})(x) = y^{\ast}(Tx) = y^{\ast}((\sum^{\infty}_{k=1} x_k,0,0,...))= y_1^{\ast}(\sum^{\infty}_{k=1} x_k)+y_2^{\ast}(0)+y_3^{\ast}(0)+...=y_1^{\ast}(\sum^{\infty}_{k=1} x_k)=y_1^{\ast}(x_1)+y_1^{\ast}(x_2)+... \end{align*}

Because $$y^{\ast}_j$$ is linear $$\forall j\in \mathbb{N}$$ $$\Rightarrow y_j^{\ast}(0)=0$$

So it appears to me that my $$T^{\ast}$$ must satisfy $$T^{\ast}(y_1,y_2,y_3,...)=(y_1,y_1,y_1,...)$$.

Is that correct? Can I write my $$T^{\ast}$$ more explicit?

• what are $y_i^*$? are you using the fact that $(\ell^1)^*\cong\ell^\infty$? Commented Jun 14, 2021 at 19:42
• I would prefer to write $$T^*(y^*)(x)= \left( \sum_{k\geq 1} x_k \right) y^*(e_1),$$ where $e_1=(1,0,\dots)$. Commented Jun 14, 2021 at 19:45
• Well I considered $y_i^{\ast}$ as coordinates of $y^{\ast}=(y_1^{\ast}, y_2^{\ast},...)$ but I probably should get rid of the '$\ast$' because it's confusing? And yes, I'm using $(l^1)^{\ast}\cong l^{\infty}$ there. Commented Jun 14, 2021 at 19:50

The adjoint of an operator $$T:X\to Y$$ between Banach spaces is the operator $$T^*:Y^*\to X^*$$ defined by $$T^*(\phi)=\phi\circ T$$. It is easy to see that $$T^*$$ is a well-defined linear operator and $$\|T^*\|=\|T\|$$ so when $$T$$ is bounded, $$T^*$$ is bounded.
In our case, we have that $$T^*:(\ell^1)^*\to(\ell^1)^*$$ defined as $$T^*(\phi)=\phi\circ T$$. Now assume that $$(x_n)\in\ell^1$$. Then $$T^*(\phi)(x_n)=\phi\circ T(x_n)=\phi(\sum_{k=1}^\infty x_k,0,0,\dots)$$
We have an isometric isomorphism $$(\ell^1)^*\cong\ell^\infty$$ which is defined by $$J:\ell^\infty\to(\ell^1)^*$$ as follows: if $$y=(y_n)\in\ell^\infty$$, then $$y\mapsto J_y$$, where $$J_y(x_n)=\sum_ny_nx_n$$.
Now for $$y=(y_n)\in\ell^\infty$$ we have that $$T^*(J_y)(x_n)=J_y(\sum_{k=1}^\infty x_k,0,0,\dots)=y_1\sum_{k=1}^\infty x_k+y_2\cdot0+y_3\cdot0+\dots=\sum_{k=1}^\infty y_1x_k=J_{z}(x_n)$$ where $$z=(y_1,y_1,y_1,\dots)\in\ell^\infty$$. This shows that $$T^*(J_y)=J_z$$, i.e. if we identify $$(\ell^1)^*$$ with $$\ell^\infty$$ through the isometric isomorphism $$J$$, then $$T^*(y_1,y_2,y_3,\dots)=(y_1,y_1,y_1,\dots)$$, as OP states.