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There may be some similar questions being asked but I didn't see anything that was quite what I was looking for so I'll ask the question here.

Let $X = (C[0,1],||.||_{\max})$ and $T:X \to X = \int^{t}_{0}x(\tau)d\tau$.

The norm is defined in the following way: for $x(t) \in C[0,1]$, $||x(t)||_{max}= \textrm{max}\{|x(t)|:t \in [0,1]\}$.

I want to show that $T$ is a bounded linear operator and find $||T||= \textrm{sup}\left\{\frac{||T(x(t))||}{||x(t)||}:x(t) \in C[0,1], x(t) \neq 0\right\}$.

Showing that $T$ is a bounded linear operator is simple but I'm not sure about finding the norm of $T$. By the Extreme Value Theorem, we know that a maximum exists for $x(t)$ and $T(x(t))$ on $[0,1]$. But the supremum of the quotient of all such values may be infinite. How do we show this is not the case?

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When we use the bound $$\tag{*}\int_0^t|x(\tau)|\mathrm{d}\tau\leqslant \lVert x\rVert_\infty,$$ there is no loss if and only if $x$ is constant so the norm is greater than $1$.

Inequality (*) for general $x$ shows that the norm is actually $1$.

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  • $\begingroup$ I'm not sure where this is coming from. The LHS is another function in $X$ while the RHS is a real number. How are you comparing them? $\endgroup$
    – inkievoyd
    Commented Oct 6, 2014 at 16:05
  • $\begingroup$ The bound holds for any $t\in [0,1]$. $\endgroup$ Commented Oct 6, 2014 at 16:06
  • $\begingroup$ So you are saying that for $t \in [0,1]$ ,$\int_{0}^{t}x(\tau)d\tau \leq \int_{0}^{t}|x(\tau)|d\tau \leq \textrm{sup}\{|x(t)|:t \in [0,1]\}$? $\endgroup$
    – inkievoyd
    Commented Oct 6, 2014 at 16:42
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ Commented Oct 6, 2014 at 17:02

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