Equivalence of norms: $\|x\|$ and $\|x\|_1=\|x\|+\lvert\,f(x)\rvert$ Let $f: X\to \mathbb{R}$ be a linear functional and $\|\cdot\|_1$ is defined as follow $\|x\|_1=\|x\|+|f(x)|$. Prove or disprove $\|\cdot\|_1$ is equivalent to $\|\cdot\|$ iff $f$ is continuous.
I've tested some examples and get the answer is positive. 

The answer is AFFIRMATIVE
First suppose that the two norms are equivalent and $a$ and $b$ be such that $a\|x\|\leq \|x\|_1\leq b\|x\|$ for all $x\in X$. Then $|f(x)|=|\|x\|_1-\|x\||\leq \|x\|_1+\|x\|\leq (1+b)\|x\|$ for all $x\in X$. Therefore, $f$ is continuous.
Conversely, let us assume that $f$ is continuous. It is clear from the definition of $\|\cdot\|_1$ that $\|x\|_1\geq \|x\|$. Also $\|x\|_1 = \|x\|+|f(x)| \leq \|x\| + \|f\|\|x\| = (1+\|f\|)\|x\|.$ Therefore, $\|x\|\leq \|x\|_1\leq (1+\|f\|)\|x\|$, which shows that the two norms are equivalent.
