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I asked a question that got a great answer referring to another question, but can't follow a line in that answer.

The question/answer that got referenced was identity operator isn't bounded .

What I can't follow is the following:

$$ \|f\|_1 = \frac{(n)(1/2n)}{2}=1 $$ but $\|f\|_\infty=n$. Consequently $$ \|T^{-1}\|=\sup_{\|f\|_1\le 1}\|Tf\|_\infty\ge\sup_{n\in\mathbb{N}}\|f_n\|_\infty=\infty, $$ so $T^{-1}$ is not bounded.

I am sure this is stupidity/ignorance on my part, but it looks to me like $$ \|f\|_1 = \frac{(n)(1/2n)}{2}=1 $$ for all $n$. So why does the answer say $\|f\|_\infty=n$?

Also, can anyone, in addition, provide an intuitive example? Perhaps this is the closest one can get, intuitively. But my simple-minded intuition on what it means for an operator to be 'bounded' makes it hard to see how you could get one that is bounded 'on the way out', but not 'on the way back'.

eSurfsnake
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    What are you asking in the first part of your question? Are you unsure why $|f|\infty=n$ in that particular example? Because that's just a straight forward application of the definition of $|\cdot |\infty$. – Demophilus Nov 05 '17 at 21:19
  • Also maybe a bit of your intuition is recovered when you go from normed spaces to Banach spaces. In Banach spaces the open mapping theorem holds, so an operator between Banach spaces that is bounded 'on the way out', as you say, will be bounded 'on the way back'. – Demophilus Nov 05 '17 at 21:22
  • Can you point me to why $|f|_\infty=n$ ? And, yes, the Banach space intuition is where I seem to be coming from. Thanks for the reference to the open mapping theorem. It had specific references to Rudin. – eSurfsnake Nov 05 '17 at 22:01
  • To write it explicitly, for $x\in [0,1/2n]$ we have $f_n(x) = n-2nx$ and for $x \in [1/2n, 1]$ we have $f_n(x) = 0$. It's pretty clear that $f_n$ is positive and decreasing, so it's largest value is equal to $f_n(0)=n$. – Demophilus Nov 05 '17 at 22:06
  • Thanks...I got stuck in what looked like the calculation of the area of the triangle being the norm, not the supremum over the range. Time to take a mental break. – eSurfsnake Nov 05 '17 at 22:11
  • Another fact that might play into your confusion, is that any two norms on finite dimensional spaces are equivalent. This has to do with the fact that the closed unit ball is compact in finite spaces but it isn't in infinite dimensional spaces. In fact there's really only one way to make a finite dimensional space into a Hausdorff topological vector space. Furthermore all linear maps between finite dimensional normed spaces, are continuous. All of this completely breaks down when you go to infinite dimensional spaces. – Demophilus Nov 05 '17 at 22:11
  • Got it. I looked at Rudin (Real & Complex Analysis, p. 65). he say @suppose $f: X\to[0,\infty] $, and suppose S is set of all real $\alpha$s.t.$\mu(f^{-1}((\alpha,\infty])=0$, – eSurfsnake Nov 06 '17 at 00:17
  • Let us continue this discussion in chat. – eSurfsnake Nov 06 '17 at 00:27

1 Answers1

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Your problem is the following. There is no fixed $f$. There is only a sequence $f_n$. And for this sequence we have $\|f_n\|_1 = 1$ and $\|f_n\|_\infty = n$. It was just a typo in John Griffin's answer.

I also wonder why you checked his answer while you still don't understand what he wrote. Or is eSurfsnake $\neq$ Newbie?

amsmath
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  • The answer got me 99% of the way there. The shortcoming in fully following it reflects my lack of knowledge, and didn't reflect on the quality of the answer...you can see the one point that, on reflection, I realized I lacked the right foundation on was a pretty technical one. – eSurfsnake Nov 05 '17 at 22:06