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 $Y$.


1 Answer 1


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 $Tx_0$.


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