Use bounded inverse theorem to show that $C[0,1]$ is incomplete in the $||.||_p$ norm for any $1\leq p<\infty$.

So we define the identity (bijective) map $I:(C[0,1],||.||_\infty)\to(C[0,1],||.||_p)$ by $f\mapsto f$ for $f\in C[0,1]$. Now $$||I(f)||_p=||f||_p\leq||f||_\infty$$ This shows that $I$ is continuous. Since $(C[0,1],||.||_\infty)$ is a well-known Banach space, so to achieve a contradiction by BIT, we need to show that $I^{-1}$ is not bounded in $||.||_\infty$ norm. But $||I^{-1}(f)||_\infty=||f||_\infty$, after that how to show $||f||_\infty\geq M$ for some $M$ I don't understand. Is any other type of operator other than identity is needed to be defined for that purpose? Any help is appreciated.


Let $f_n:[0,1]\ni x\mapsto x^n$. Then, $$||f_n||_p=\bigg(\int_0^1x^{np} dx\bigg)^{1/p}=\frac{1}{\sqrt[p]{np+1}}$$ Hence, $g_n:=\sqrt[p]{np+1}f_n$ has the property that $||g_n||_p=1$. But, note that $||g_n||_\infty=\sqrt[p]{np+1}\to \infty$ as $n\to \infty$.

Now, $||I^{-1}||=\sup\big\{||I^{-1}(g)||_\infty:g\in C[0,1],||g||_p\leq 1\big\}=\infty.$ In other words, $I^{-1}$ is not bounded linear operator.

So, if $\big(C[0,1],||•||_p\big)$ were complete then by open mapping theorem $I$ would be an open map, i.e. $I^{-1}$ would be continuous map, which is impossible.


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