Here's Theorem 4.3-3 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $X$ be a normed space and let $x_0 \neq 0$ be any element of $X$. Then there exists a bounded linear functional $f_0$ on $X$ such that $$ \Vert f_0 \Vert = 1 \ \ \ \mbox{ and } \ \ \ f_0(x_0) = \Vert x_0 \Vert.$$

Let $X$ and $Y$ be normed spaces, both real or both complex, and let $T \colon X \to Y$ be a bounded linear operator. Let $X^\prime$ and $Y^\prime$ denote, respectively, the dual spaces (i.e. spaces of all the bounded linear functionals) of $X$ and $Y$. Then the adjoint operator of $T^\times$ of $T$ is defined to be the operator $T^\times \colon Y^\prime \to X^\prime$ as follows: for each $g \in Y^\prime$, we define $T^\times g \colon= f \in X^*$, where $X^*$ denotes the space of all the linear functionals defined on $X$ and $$f(x) \colon= g(Tx) \ \mbox{ for each } \ x \in X.$$

Now Theorem 4.5-2 in Kreyszig states

The adjoint operator $T^\times $ of a bounded linear operator is linear and bounded with $$\Vert T^\times \Vert = \Vert T \Vert.$$

It is clear to me how $T^\times$ is linear and bounded with $$\Vert T^\times \Vert \leq \Vert T \Vert.$$

Now Kreyszig starts off the proof that $\Vert T^\times \Vert \geq \Vert T \Vert$ as follows:

Theorem 4.3-3 implies that for every nonzero $x_0 \in X$ there is a $g_0 \in Y^\prime$ such that $$\Vert g_0 \Vert = 1 \ \mbox{ and } \ g_0( T x_0 ) = \Vert T x_0 \Vert.$$

This is evident if $T x_0$ is non-zero.

But how does the last conclusion hold if $x_0 \in X$ is non-zero, but $T x_0$ equals the zero vector in $Y$?

Of course, Theorem 4.3-3 applies only to non-zero elements of $Y$.


If $T=0$, then the adjoint $T^\times$ is also equal to zero and the inequality is trivial. If $T \neq 0$, there exists $x_0$ with $Tx_0 \neq 0$. Theorem 4.3.3 is then applied to $Tx_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.