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. 
 A: If $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$, then it converges in $\|\cdot\|$ because
$$
               \|x-x_n\| \le \|x-x_n\|_1.
$$
Conversely, if $\{ x_n \}$ converges to $x$ in $\|\cdot\|$, then $\{f(x_n)\}$ converges to $f(x)$ because $f$ is continuous; therefore, $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$.
A: Hint. It suffices to prove the following:
$$
\|x_n-x\|\to 0 \quad\text{iff}\quad \|x_n-x\|_1\to 0.
$$
A: 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.
A: 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.
A: Let us denote two norms as $\mu(x) = \|x\|_1 $ and $\nu(x) = \|x\|$. 
From the definition of  $\|x\|_1 $  we have $\mu(x) = \nu(x) + |f(x)|$.
Two norms $\mu$ and $\nu$ are equivalent if there exist two constants $c,C\in\mathbb{R}$ such that $\forall x\in X$  $c\nu(x) \leq \mu(x) \leq C\nu(x)$.
First, since  $\mu(x) = \nu(x) + |f(x)|$ and $|f(x)|>0$ we conclude that $\nu(x) \leq \mu(x)$.
Second, $f$ is a linear functional, so it is bounded on $X$, i.e. $\forall x \in X \ |f(x)|\leq M\|x\|$, where $M$ is some non-negative constant. But then 
$$\mu(x) = \nu(x) + |f(x)| \leq \nu(x) + M\|x\| = \nu(x) + M\nu(x) = \big(1+M\big)\nu(x) $$
Therefore we have 
$$ \mu(x)\leq \big(1+M\big)\nu(x) $$
Finally, we write
$$\nu(x) \leq \mu(x) \leq \big(1+M\big)\nu(x),$$
i.e.   norms $\mu$ and $\nu$ are equvalent.
Q.E.D.
