Why is boundedness different with an infinite-dimensional space? 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'.  

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?
