# An application of bounded inverse theorem

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.

## 1 Answer

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.