I have an exercise that I need to solve and I can't finish it. Let $k \in \mathcal{C}([0,1] \to \mathbb{R})$. Proove that this operator : $$ \begin{array}{ccccc} T & : & \left(\mathcal{C}([0,1] \to \mathbb{R}), ||.||_\infty \right) &\rightarrow & \mathbb{R}\\ & & f &\mapsto & \int_0^1 k(t)f(t)dt\\ \end{array} $$ is continuous, calculate its norm.

As this is a linear operator, I've shown that it is bounded by $||k||_1$, which prooves that it's continuous. Therefore $|||T||| \leq ||k||_1$. I think that, indeed, $|||T||| = ||k||_1$.

But, I have a problem finding a function $g$ such as $||g||_\infty \leq 1$ and $|T(g)| = ||k||_1$. I thought about $g(x) = sign(k)$, with this function that would work. Problem : the function $g$ is not continuous… Solution : find a sequence of continuous functions $g_n$ which, starting a certain rank, verifies $\forall n, \ ||g_n||_\infty \leq 1$ and has for limit $g$. Question : does such a sequence exist ? If yes, what is its expression ?

Thanks in advance and sorry for my bad english.


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