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Suppose $X$ and $Y$ are normed spaces over $\mathbb{R}$ and suppose $T: X \rightarrow Y$ is a bounded linear map. I want to prove that the adjoint map $T^\star : Y^\star \rightarrow X^\star$ is bounded as a linear map also, but I get super confused when trying to deal with the norms on the dual spaces since there are four norms, to deal with and I am unsure whether it is the operator norm of $T^\star$ or the dual norms i need to take?

I suppose the question is that since $\Vert T(x) \Vert_X \leq M \Vert x \Vert_Y$ it should be possible to find a $K$ s.t. $\Vert T^\star(l) \Vert_X^{\star} \leq K \Vert l \Vert_Y^{\star}$, but expanding these seems a little unfeasible.

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Let $A \in L(X,Y)$, $\tilde y \in Y^*$, $x \in X$. Then from $\left(A^*(\tilde y)\right)(x) := \tilde y (A(x))$ you get

$$|A^*(\tilde y)(x)|= |\tilde y (A(x))|≤\|\tilde y\|_{Y^*} \|A(x)\|_Y ≤ \|\tilde y\|_{Y^*} \|A\|_{L(X,Y)} \|x\|_X$$

For $A^*$ to be bounded you must consider the following expression

$$\|A^*\|_{L(Y^*,X^*)}=\sup_{\tilde y \in Y^*, \ \|\tilde y\|_{Y^*}≤1} \left\{\|A^*(\tilde y)\|_{X^*}\right\}$$

Note that you have the following inequality, which follows from the first one

$$\|A^*(\tilde y)\|_{X^*}=\sup_{x \in X, \ \|x\|_X≤1}\{|A^*(\tilde y)(x)|\}≤\sup_{x \in X, \ \|x\|_X≤1}\{\|\tilde y\|_{Y^*} \|A\|_{L(X,Y)}\ \|x\|_X \}$$

Putting it together gives you

$$\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$$

Note further that the construction gives you $\|A^{**}\|_{L(X^{**},Y^{**})}≤\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$. If you restrict $A^{**}$ to the subspace given by the embedding of $X$ into $X^{**}$ you get precisely $A$ again (composed with the embedding of $Y$ into $Y^{**}$). So $\|A^{**}\|_{L(X^{**},Y^{**})}≥\|A\|_{L(X,Y)}$ also holds, and the two inequalities are equalities, so also $\|A^*\|=\|A\|$.

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  • $\begingroup$ I'm not sure of the adjoint of an operator $X \to Y$ (instead of $X \to X$) are you saying that sometimes $\|A^*\| \ne \|A\|$ ? it is when $Y$ is not isomorphic to $X$ ? $\endgroup$ – reuns Apr 24 '16 at 14:49
  • $\begingroup$ I have edited the answer to show $\|A^*\|=\|A\|$ in general. $\endgroup$ – s.harp Apr 24 '16 at 14:54
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the definition of the adjoint of a bounded operator $T : X \to X$ is

$$ \text{for every }\ x \in X, y \in X^* : \qquad y(T(x)) = (T^* y)(x)$$ then note that for any $x \in X$ :$$\|x\|_{X} = \max_{y \in X^*, \ \|y\|_{X^*} =1} |y(x)|$$ hence $$\|T\| = \max_{\|x\|_X=1} \|Tx\| = \max_{\|x\|_X=1}\max_{\|y\|_{X^*}=1} |y(Tx)| = \max_{\|y\|_{X^*}=1} \max_{\|x\|_X=1}|y(Tx)|$$ $$ = \max_{\|y\|_{X^*}=1} \max_{\|x\|_X=1}|(T^*y)(x)| = \max_{\|y\|_{X^*}=1} \|T^*y\|_{X^*} = \|T^*\|$$

(this can easily be extended to the case $T : X \to Y$, if the adjoint exists)

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  • $\begingroup$ (and when the normed space is not a Banach space, replace $\max$ by $\sup$ because the maximiser is in the Banach space, not always in the incomplete normed vector space) $\endgroup$ – reuns Apr 24 '16 at 14:54

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