Linear functional $f$ is continuous at $x_0=0$ if and only if $f$ is continuous $\forall x\in X$? Let $f$ be a linear functional on a normed space $(X, \|\cdot\|)$. Prove that $f$ is continuous at $x_0=0$ if and only if $f$ is continuous at every $x\in X$.
I understand that the $\Leftarrow$ is trivial but what about the other way?
 A: Suppose $f$ is continuous at some $x_0\in X$ and let $x_n\to x^*$, for some $x^*\in X$. Then the sequence $x_n-x^*+x_0\to x_0$ and thus $\|f(x_n)-f(x^*)\|=\|f(x_n-x^*+x_0)-f(x_0)\|\to 0$. Hence $f$ is continuous at $x^*$ as well.
A: Consider
$$\lVert T(x-x_0)\rVert=\lVert T(x)-T(x_0)\rVert$$
So if $\lVert x-x_0\rVert<\delta$ we have that $\lVert T(x)-T(x_0)\rVert=\lVert T(x-x_0)\rVert<\epsilon$ by continuity at $0$.
A: if $f$ be linear function on a normed space $(X,\| \cdot\|)$ then these conditions are equivalent.
$(1)$ : $\|f\| < \infty  $
$(2)$ : $f$ is uniformly continuous
$(3)$ :  $f$ is continuous
$(4)$ : $f$ is  continuous in $0$
$(1)$ $\to$ $(2)$ : $\forall x,y \in X$ and $\forall \epsilon> 0 $ $\|f(x)-f(y)\|$= $\|f(x-y)\|$ $\leq$ $\|f\|$ $\|x-y\|$,  then if $\delta$ = $\frac{\| \epsilon\|}{\|f\| +1}$ we claim that $f$ is uniformly continuous
$(2)$ $\to$ $(3)$ : obvious
$(3)$ $\to$ $(4)$ : obvious
$(4)$ $\to$ $(1)$ : let $f$ be continuous in $0$. the there exist $\epsilon = 1$ and $\delta > 0$ such that if $\| u \|< \delta$ then $\|f(u)\|< 1$
$\|\frac{x}{\frac{2}{\delta}\|x\|}\|$ < $\delta$ $\longrightarrow$ $\frac{\|f(x)\|}{\|x\|}$= $\frac{\|f(u)\|}{\|u\|}$ < $\frac{1}{\frac{2}{\delta}}$= $\frac{\delta}{2}$ $\longrightarrow$ $\|f\|$ < $\infty$ $\quad$
A: Proof.
$\implies$ obviously
$\Longleftarrow$ 
Assume L is continuous at $x_0 \in E$. We define sequence $x_1,x_2,\dotsm \in E$  and for any point $x\in E$  such that $||x_n-x ||\rightarrow 0$.
Since L is linear mapping, we have
$$
||L(x_n-x+x_0-x_0)||=||L(x_n-x+x_0)-L(x_0)||
$$
Since L is continuous at $x_0\in E$, for every sequence converging to $x_0$, the $L$ converges $L(x_0)$. We have 
$$
||L(x_n-x+x_0)-L(x_0)||\rightarrow 0
$$
Q.E.D.
