Question: For example 

[$T:C[0,1]\to C[0,1]$ defined by $T(f)(t)=\frac{t}{1+t^2}f(t)$ find norm of $T$](https://math.stackexchange.com/questions/3711725/tc0-1-to-c0-1-defined-by-tft-fract1t2ft-find-norm-of-t?noredirect=1#comment7628283_3711725)

I first find $||T||\le A$ and then $||T||\ge B$ for a spesific element in the domain of $T$ these two equalities surely says that norm of $T$ is bounded but how can one be sure about there exist a norm between $A$ and $B$ since I am new in learning functional analysis and it sounds stupid but I cannot make a logical chain to link that there must be some $||T||$. If there is one I can show that it is unique but how to sure about existence and 

**Is there an example that T is linear from one normed space to another one with no norm?**

Answer: 
Operator norms are defined as infima of sets.  

$$ ||A|| = \inf \{ c \geq 0 : ||A v|| \leq c||v|| \text{ for all } v \in V \} $$
If you have shown that a nonempty set is bounded, you have shown that its infimum exists and is finite.
The norm of a linear map can be $\infty$. In fact it is well known that every infinite dimensional space supports an unbounded linear functional. If $F$ is such a functional on $C[0,1]$ then we can regard $F$ as a linear map from $C[0,1]$ into itself since any real number can be treated as a constant function. This gives a linear map on $C[0,1]$ with norm infinity. 

Computation of the norm of you operator $T$

Recall that $\frac t {1+t^{2}} \leq \frac 1 2$. This comes from the fact that $(t-1)^{2} \geq 0$. Hence $\|\frac t {1+t^{2}} f(t)\| \leq \frac 1 2 \|f\|$ and this shows that $\|T\| \leq \frac 1 2$. Now let $f(x)=x$ for all $x$. The $\|f\|=1$ and $\|Tf\|$ is the supremum of $\frac t {1+t^{2}}$ over $[0,1]$. We already know that the supremum is $\leq \frac 1 2$. The value $\frac 1 2$ is attained when $t=1$ so $\|Tf\|=\frac 1 2$. It follows from the definition of norm of an operator that $\|T|| \geq \frac 1 2$. Hence we have $\|T|| = \frac 1 2$.