Space bounded from norm real analysis

Question

Question:

\begin{equation} \text{Let} \: \: \: T : (C[0, 1], \|\cdot\|_{\infty}) \rightarrow \mathbb{R} \: \: \: \text{with} \: \: \: T(f) = \int_0^1 f(t)\,dt. \end{equation} Prove that $T$ is bounded and calculate its norm.

First thought: To show that $T$ is bounded, I need to find a number $K$ such that $\|Tf\| \leq K \cdot \|f\|$ for all $f$ in $C[0, 1]$. I have: \begin{equation} \|Tf\| = |T(f)| = \left|\int_0^1 f(t)dt\right| \leq \int_0^1 |f(t)|dt \leq \int_0^1 \|f\|_{\infty} dt = \|f\|_{\infty} \end{equation}

So, $T$ is bounded with norm at most $1$. I need to calculate the norm from here right?

Answer 1

Your work is correct. Now consider any constant, nonzero function: $T(c) = c = 1\cdot c$, so it attains the upper bound.

Answer 2

Yes!. At this point you know that $$||T|| = sup_{x \in C([0,1])} \frac{|T(f)|}{||f||_{\infty}}\leq 1$$ so if you prove that there exist an $x \in C([0,1])$ such that $\frac{|T(x)|}{||x||_{\infty}} = 1$, then it follows that ||T|| = 1. Hint:$f \equiv 1 \in C([0,1])$