Let $f : \langle V_1, \|\cdot\|_1\rangle \to \langle V_2, \|\cdot\|_2\rangle$ be linear. Then if $f$ is continuous at some $v \in V_1$, then it is continuous on all of $V_1$. Without appealing to boundedness, because that's an equivalent condition that I'm proving in my next problem.
My attempt. Proof:
We want to show that for all $v \in V, \ \ v_n \to v \implies f(v_n) \to f(v)$. Assume we know that $f$ is continuous at $0$. then $v_n \to v \implies v_n - v \to 0 \implies f(v_n - v) \to 0 \implies f(v_n) \to f(v)$ by linearity.
But I'm having trouble proving that given any single point of continuity $v \in V$, $f$ is also continuous at $0$.
My best guess is: $v_n \to v $ and $ w_n \to 0 \implies v_n - w_n \to v \implies f(v_n - w_n) = f(v_n) - f(w_n) \to f(v) \implies f(w_n) \to 0$
