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I have a little trouble understanding adjoint operators in spaces without inner products. So the definition of the adjoint operator is the following:

Definition (Adjoint operator $T^\times$): Let $T: X \to Y$ be a bounded linear operator between normed spaces. Then the adjoint operator $T^\times: Y' \to X'$ is defined by $$(T^\times g)(x) = g(Tx) $$ where $g \in Y'$.

So I wish to find some specific adjoint operators. In the examples we have that $1 \leq p < \infty$.

Example 1. Let $a_n$ be a bounded sequence with $\lim_{n\to\infty} a_n = 0$. Then find the adjoint of the operator $T_1: \ell^p \to \ell^p$ defined by $T_1(x_n)_{n=1}^\infty = (a_n x_n)_{n=1}^\infty.$

Example 2. Let $(a,b)$ be a non-empty interval. Find the adjoint operator of $T_2:L^p(a,b)\to L^p(a,b)$ defined by $$ T_2x(t) = \int_a^b e^{t-s} x(s)\,\mathrm{d}s. $$

I know that if $p^{-1} + q^{-1} = 1$ then $(\ell^p)' = \ell^q$ and $(L^p(a,b))' = L^q(a,b)$ which I believe one should use at some point.

I don't necessarily need everything solved for me, but if someone could show the process once, maybe I could solve the rest myself.

Thanks for any help.

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Note that the isomorphism $S \colon \ell^q \to (\ell^p)'$ you know is given by $$ (Sx)(y) = \sum_n x_n y_n, \qquad x \in \ell^q, y \in \ell^p $$ Usually one omits $S$ in notation, that is identifies $\ell^q = (\ell^p)'$. For $T_1$ we have $y \in \ell^q$, $x \in \ell^p$ \begin{align*} (T_1^\times y)(x) &= y(Tx)\\ &= \sum_n y_n (Tx)_n\\ &= \sum_n y_n a_n x_n\\ &= \left(( a_n y_n)_n \right)(x) \end{align*} That is $$ T_1^\times y = (a_n y_n)_n $$ For $T_2$ note that the isomorphism is given by $$ (Sf)(g) = \int_a^b f(s)g(s)\, ds, \qquad f \in L^q, g \in L^p $$ we have for $g \in L^q$, $f \in L^p$: \begin{align*} (T_2^\times g)(f) &= g(Tf)\\ &= \int_a^b g(t)(Tf)(t)\, dt\\ &= \int_a^b \int_a^b g(t)e^{t-s}f(s)\, ds\,dt\\ &= \int_a^b \int_a^b e^{t-s}g(t)\, dt\, f(s)\,ds &= \int_a^b (T_2^\times g)(s)\, f(s)\, ds \end{align*} Hence $$ (T_2^\times g)(s) = \int_a^b e^{t-s}\, g(t)\, dt $$

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  • $\begingroup$ Thanks a lot. I'm reading your answer slowly to see if I can understand everything. $\endgroup$ – Eff Feb 17 '16 at 12:54
  • $\begingroup$ I think the use of the isomorphisms was what I needed. Thanks again! $\endgroup$ – Eff Feb 17 '16 at 14:02

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