Let $X = Y = C([0,1])$ be the space of continuous functions $f:[0,1]\to \mathbb{R}$ equipped with the norms $$ \|f\|_X := \sup_{0\le t \le 1} |f(t)|, \quad \quad \|f\|_Y := \sqrt{\int_0^1|f(t)|^2 dt}. $$ Here $X$ is a Banach space and $Y$ is a not, that is, $X$ is complete and $Y$ is not. I have read that the identity map $I:X\to Y$ is a bijective bounded linear operator in this case, but it has an unbounded inverse. I recognise that due to the lack of completeness we end up with unbounded inverses, in a general sense, but how can it be shown specifically in this case that the inverse is bounded?

  • $\begingroup$ This is better referred to as injection rather than identity. $\endgroup$ – Ranc Jul 29 '17 at 10:02
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    $\begingroup$ My edit was to change $\|f(t)\|^2$ to $|f(t)|^2$ in the def'n of $\|f\|_Y$. $\endgroup$ – DanielWainfleet Jul 29 '17 at 21:22
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    $\begingroup$ BTW. I have seen an example of an incomplete normed linear space $X$ with a bounded linear bijection $\psi:X \to X$ whose inverse is unbounded. $\endgroup$ – DanielWainfleet Jul 29 '17 at 21:50

Define, for each $n\in\mathbb N$,$$f_n(x)=\begin{cases}\sqrt{n-n^2x}&\text{ if }0\leqslant x\leqslant\frac1n\\0&\text{ if }\frac1n\leqslant x\leqslant 1.\end{cases}$$Then$$(\forall n\in\mathbb{N}):\|f_n\|_Y=\int_0^{\frac1n}n-n^2x\,\mathrm dx=\frac12.$$Therefore $\{f_n\,|\,n\in\mathbb{N}\}$ is a bounded set with respect to the $\|\cdot\|_Y$ norm. But not with respect to the $\|\cdot\|_X$ norm, since$$(\forall n\in\mathbb{N}):\|f_n\|_X=n.$$So, the identity map is not bounded.

  • $\begingroup$ Any sequence $(f_n)_n$ of members of $C[0,1]$ where $\|f_n\|_Y\to 0$ but $\|f_n||_X=1$ will suffice.... E.g. $f_n(t)=t^n.$ $\endgroup$ – DanielWainfleet Jul 29 '17 at 21:45
  • $\begingroup$ @DanielWainfleet You are entirely right. In fact, so right that I don't understand why you wrote this as a comment to my answer. It would be more natural to be a comment to the original question or, better still, to be an answer. Furthermore, it is easier to understand than my answer. $\endgroup$ – José Carlos Santos Jul 29 '17 at 21:48
  • $\begingroup$ Your answer was, I thought, good enough. $\endgroup$ – DanielWainfleet Jul 29 '17 at 21:53
  • $\begingroup$ @DanielWainfleet That's my opinion, too. But, as I wrote, your answer is easier to understand. $\endgroup$ – José Carlos Santos Jul 29 '17 at 21:54

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