While reading up on Sturm-Liouville system theory, I came across something I didn't fully understand. At one point, in the midst of proving the existence of solutions to the Sturm-Liouvill problem, the book defines the norm of the operator $T$
$$||T|| = \sup \{ ||Tu|| : u \in C([a,b]), ||u|| = 1 \}$$
where $T$ is the operator
$$(Tf)(x) = \int_a^b G(x, \xi)f(\xi)\, d\xi$$
where $G$ is the Green's function for the Sturm-Liouville system:
$$\frac{d}{dx} \left(p \frac{dy}{dx} \right) + ry + \lambda w y = 0$$
The book then states that
$$||Tu|| = \left| \left| T\left( \frac{u}{||u||}\right) \right| \right|||u|| \leq ||T|| ||u||$$
However, I don't see how the inequality follows. Perhaps I am missing some knowledge about the norm of a function, but I can't figure out why this inequality must hold for all $u \in C([a,b])$ as the book says. Any help would be much appreciated!
This is a general inequality concerning bounded linear operators between normed spaces. The Sturm-Liouville setting is actually irrelevant here.
Let $X,Y$, be normed spaces, and let $T:X\to Y$ be linear and bounded. That is, the set$$\{Tx|\|x\|=1\}\subset Y$$is bounded. The operator norm of $T$ is defined by$$\|T\|=\sup_{\|x=1\|}\|T(x)\|,$$and the inequality in question holds for any such $T$. To prove it, let $x_0\in X$. By linearity of $T$ and the norm being homogeneous, we have$$\sup_{\|x\|=\|x_0\|}\|T(x)\|=\|x_0\|\cdot\sup_{\|x\|=1}\|T(x)\|=\|x_0\|\cdot\|T\|,$$and hence$$\|Tx_0\|\leq\|T\|\cdot\|x_0\|.$$