$T$ is continuous on $X$ iff there exists a $C>0$ such that $||Tx||\le C||x||$ Suppose that $X, Y$ are normed linear space and that $T:X\to Y$ is a linear map, then prove that
a) $T$ is continuous on $X$ iff  $T$ is continuous at $0$.
and using a), prove that $T$ is continuous on $X$ iff there exists a $C>0$ such that $||Tx||\le C||x||$ for all $x\in X$.
I tried to prove it as follows:
a) ($\Rightarrow$) is straightforward.
($\Leftarrow$) Let $x\in X$ be arbitrary. Let $x_n\to x$. Suppose that $r_n:=x_n-x$. Clearly, $r_n\to 0$. $Tr_n+Tx=Tx_n$. By continuity of $T$ at $0$, $Tr_n\to 0$; and therefore, it follows that $Tx_n\to Tx$. Since $x$ is arbitrary, it follows that $T$ is continuous on $X$. This completes the proof of part $(a)$.
The second part is where I get stuck.
($\Leftarrow$) This direction is straightforward.
$(\Rightarrow)$ By a), $T$ is continuous at $0$ so there is a $\delta>0$ such that $||Tx||\lt 1$ for all $x: ||x||<\delta$. I don't know how to proceed from here to introduce $C$. Any hints on this? Thanks.
 A: As you correctly pointed out, there exists a $\delta>0$, s.t. if $||x||<\delta$ we have $||Tx||<1$.
Let $y\in X$, with $y\ne 0$. Now if we define $x:=\frac{\delta}{2}\frac{y}{||y||}$ then of course $||Tx||<1$, since $||x||<\delta$.
However: $||Ty||=||\frac{2||y||}{\delta}Tx||$ and from that on I'm sure you'll get it yourself :)
Edit: Sorry haven't seen that is has already been answered.
A: Choose $x_0\in X$ .
$\|Tx-Tx_0\|=\|T(x-x_0)\|$
Since $T$ is continuous at $0$,
$\|(x-x_0)\|<\delta$ implies $\|T(x-x_0)\|<\epsilon$

Suppose $\forall C>0, \exists x\in X$ such that $\|Tx\|\ge C\|x\|$
Hence $\forall n\in\Bbb{N}, \exists( x_n) \subset X$ such that $\|Tx_n\|\ge n\|x_n\|$
Then $y_n=\frac{x_n}{n\|x_n\|}\to 0$ but $\|Ty_n\| $ doesn't converge to $0$ which contradict the continuity of $T$ at $0$.

Infact if $T$ is a bounded linear map $\|Tx\|\le \|T\|_{op}\|x\|$
Where $\|T\|_{op}=\sup\{\|Tx\|: \|x\|\le 1\}$
A: Contrary to your claim, suppose that for every positive integer $n$ there exists $x_n \in X$ such that $\|Tx_n\| > n\|x_n\|$. Since $T$ is linear, we may assume that $\|x_n\| = 1$ for every $n$ (if not, just take $x_n' = x_n/\|x_n\|$). Hence,
$$
\|T \frac{x_n}{\sqrt{n}}\| > \sqrt{n}
$$
for every $n$, which contradicts the continuity of $T$ at $0$.
