Evaluating the norm of $\mathfrak I : E \to E$ [duplicate]

I'm having some trouble with the following exercise:

Let $$E=\mathcal C([a,b])$$ be the set of continuous maps $$f:[a,b]\to \mathbb R$$ with the supremum norm. Let $$x\in E$$ and define $$\mathfrak{I}x\in E$$ as $$(\mathfrak{I}x)(t)=\int_a^tx(s)ds$$

Prove that:

1. $$\mathfrak I\in L(E)$$ (The set of linear continuous maps from $$E$$ to itself with the norm $$\|f\|=\sup\limits_{\|x\|\leq1} \|f(x)\|$$) and evaluate $$\|\mathfrak I\|$$.

2. Show that $$\mathfrak I$$ has no eigenvalues.

I was able to prove (2) and in the first one, I managed to prove that $$\mathfrak I \in L(E)$$ and although I was able to find an upper bound to $$\|\mathfrak I\|$$, I wasn't able to evaluate its precise value.

How can this be done?

You can bound: $$\left|\int_a^t x\right|\le(t-a)\|x\|\le(b-a)\|x\|$$
So, the norm of your operator is less than or equal to $$b-a$$. But, as Geetha mentions, taking $$x$$ to be a nonzero constant function shows the norm is precisely $$(b-a)$$.
That’s because if an operator $$T$$ has $$\|T\|\le N$$ and $$\|Tx\|=N\|x\|$$ for some $$x$$, then $$\|T\|=N$$. This is a straightforward fact.
A more general fact holds. Assume a linear operator $$T: C[a,b]\to C[a,b]$$ is positive, i.e. for $$f\ge 0$$ we have $$Tf\ge 0.$$ Then $$T$$ is bounded and $$\|T\|=\|T1\|_\infty.$$ Indeed for any $$f\in C[a,b]$$ we have $$\|f\|_\infty 1\pm f\ge 0.$$ Hence $$\|f\|_\infty T1\pm Tf=T(\|f\|_\infty 1\pm f)\ge 0$$ Thus $$|Tf|\le \|f\|_\infty T1$$ which implies $$\|Tf\|_\infty \le \|T1\|_\infty \|f\|_\infty\ {\rm and }\ \|T\|\le \|T1\|_\infty$$ The converse inequality $$\|T\|\ge \|T1\|_\infty$$ holds obviously, which means the norm is attained at $$f\equiv 1.$$