Is $C[a,b]$ complete when given the norm $||f||:=\sup_{x\in [a,b]} \Big|\int_a^xf(t)dt \Big|$?

Question

Is $C[a,b]$ complete when given the norm $||f||:=\sup_{x\in [a,b]} \Big|\int_a^xf(t)dt \Big|$ ?

Answer 1

Let $X$ denote $C([a,b])$ with your proposed norm, and let $Y$ denote $C([a,b])$ with the uniform norm. Consider the map $T : X \to Y$ defined by $(Tf)(x) = \int_a^x f(t)\,dt$. Then $T$ is an isometry. If $X$ is complete, then its image $TX \subset Y$ is also complete, hence must be a closed subspace of $Y$. But $TX$ consists precisely of the $C^1$ functions (satisfying $F(a)=0$), which are not closed in the uniform norm on $C([a,b])$.

Alternate proof: consider the identity map $I : Y \to X$. Since $\|f\|_X \le (b-a) \|f\|_Y$, the map $I$ is continuous. It is also bijective. Clearly $Y$ is complete, so if $X$ were also complete, the open mapping theorem would say that $I^{-1}$ is continuous, meaning the $X$ and $Y$ norms are equivalent. But this is clearly not true; consider for example a sequence $f_n$ with narrow bumps of height 1. They converge to 0 in the $X$ norm but not in the $Y$ norm.