Adjoint of a Bounded Linear Operator – Boundedness Proof

functional-analysislinear algebra

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.

Best Answer

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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