1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ This is better referred to as injection rather than identity. $\endgroup$ – Ranc Jul 29 '17 at 10:02
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.