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. 

It is clear that $\|x\| \le \|x\|_1$.
If $f$ is continuous, it is a bounded operator and so there is some $M$ such that $|f(x)| \le M\|x\|$. Then
$\|x\|_1 \le (1+M) \|x\|$, and so $\|\cdot\|_1$ and $\|\cdot\|$ are equivalent.
If the two norms are equivalent, there is some $L$ such that
$\|x\|_1 \le L \|x\|$, and so
$|f(x)| \le (L-1) \|x\|$. Hence $f$ is bounded and so continuous.
