nonzero bounded linear operator is not bounded on $X$ Let $X$ and $Y$ be normed vector spaces. 
    A function $f: X \rightarrow Y$ is said to be bounded on $X$ if there exists a $M>0$ such that $\| f(x) \| \leq M$ for all $x \in X.$
    Show that a nonzero bounded linear operator $T:X \rightarrow Y$ is not bounded on $X.$
My attempt: I prove by contraposition. 
Suppose that $T$ is bounded on $X.$ So there exists $M^{\prime}>0$ such that for all $x \in X,$ we have $\| T(x) \| \leq M^{\prime}.$
I need to prove that for all $M>0,$ there exists $x \in X$ such that $\| x \| \leq 1$ but $\| T(x) \| > M.$
Let $M>0$ be given. If $M>M^{\prime},$ then there does not exist $x \in X$ such that $\| T(x) \| >M$ as $M^{\prime}$ is an upper bounded for $\| T(x) \|$. 
I do not know whether I am in the right track or not. It seems to me that question is wrong. 
 A: Hint: take $x \in X$ such that $T(x) \neq 0$, Now Note that $T(\lambda x) =\lambda T(x)$  for all $\lambda \in R,  $.
BTW you are not in right track. when you say proof by contraposition,you must show that :  $T$ is bounded implies $T=0.$
However I suggest you prove it directly, i.e., show that if $T\neq 0$ implies $T$ is unbounded. 
A: You can also use contraposition. You have to show:
if $T$ is bounded on $X$, then $T$ is not a non-zero, bounded, linear operator. Or, to put it in other words,
you have to check that


*

*$T$ is zero, or

*$T$ is not bounded on $B_X$ or

*$T$ is not linear.


Let us start with $\|T(x)\| \le M'$. If $T$ is not linear, we are done. Hence, suppose that $T$ is linear. Now, let $x \in X$ be arbitrary. Thus, $\lambda \, \|T x \| = \| T(\lambda \, x)\| \le M'$ holds for all $\lambda > 0$. Thus, $T x = 0$, i.e., $T = 0$.
Note that you cannot show that $T$ is not bounded on $B_X$ (since that is trivially satisfied). Indeed, one has the stronger statement: A non-zero, linear operator is not bounded on $X$.
