There is no norm $\| \cdot \|$ and constant $C$ such that $C^{-1} \| f \| \le \| f \|_{L^1_w} \le C\| f \|$ in weak $L^1$ space $L^1_w$ Define a Weak $L^1$ Space $L^1_w$ as following. Let $f$ is measurable function and define $\lambda_f (t) = m({x:\vert f(x) \vert > t})$. And define
$$\| f \|_{L^1_w} = \sup\limits_{t>0} (t\lambda_f (t))$$
Then $f \in L^1_w$ iff $\| f \|_{L^1_w} < \infty$.
Then how to prove that there is no norm $\| \cdot \|$ on $L^1_w$ and a constant $C$ such that 
$$C^{-1}\| f\| \le \| f \|_{L^1_w} \le C\| f\| , \forall f \in L^1_w$$
Here are my idea. It's easy to see that $C$ must be equal or greater than $1$. And there is a weak triangle inequality for weak norm $\| \cdot \|_{L^1_w}$
$$\| f + g \|_{L^1_w} \le 2(\| f \|_{L^1_w} + \| g \|_{L^1_w})$$
For equality holds, there is an example. Let $f,g$ supported on $[0,1]$ and $f = x,g=1-x$. Then $\| f+g \|_{L^1_w} = 1, \| f \|_{L^1_w} = \|g \|_{L^1_w} = \frac{1}{4}$.
If we can decompose every $f$ to $h_1 + h_2$ such that $\|f \|_{L^1_w} = 2(\| h_1 \|_{L^1_w} + \| h_2 \|_{L^1_w})$, then we have 
$$\|f \|_{L^1_w} = 2(\| h_1 \|_{L^1_w} + \| h_2 \|_{L^1_w}) = \cdots = \sum\limits_{i=1}^{2^n} 2^n ( \| h_{n,i} \|_{L^1_w}) \ge C^{-1}\sum\limits_{i=1}^{2^n} 2^n ( \| h_{n,i} \|)$$
where $\sum\limits_{i=1}^{2^n} h_{n,i} = f$. And also we have
$$\| f \|_{L^1_w} \le C \| f \| \le C\sum\limits_{i=1}^{2^n} \| h_{n,i} \|$$
If $2^n > C^2$ this would lead to a contradiction.
 A: Not yet a full answer. Interesting question that I wanted to know the answer to for some time. I googled and found this PDF.
There, ~killip says the impossibility (at least for dimension 1, though I imagine a similar result in higher dimensions through radial functions) can be deduced from noticing that (writing $\smash{\|\|^*_{L^1_w}}$ instead of $\smash{\|\|_{L^1_w}}$ since it is not a norm)
$$ \left\|\sum_{k=1}^N \frac1{|x-k|}\right\|^*_{L^1_w} \sim N\log N,\qquad \sum_{k=1}^N\left\| \frac1{|x-k|}\right\|^*_{L^1_w} \sim N$$
So if $\|\bullet\|$ was somehow norm equivalent to $\|\bullet\|_{L^1_w}^*$, we would have
$$    N \log N \lesssim \left\|\sum_{k=1}^N \frac1{|x-k|}\right\|^*_{L^1_w} \lesssim \left\|\sum_{k=1}^N \frac1{|x-k|}\right\| \lesssim   \sum_{k=1}^N   \left\|\frac1{|x-k|}\right\|  \lesssim  \sum_{k=1}^N\left\| \frac1{|x-k|}\right\|^*_{L^1_w} \lesssim N,  $$
which is impossible when $N$ is sufficiently large that the implied constants mean nothing.
Now, about the two identities - The right identity is obvious since $\left\|\frac1{|x-k|}\right\|^*_{L^1_w} = 2$, but I haven't had any luck verifying the lower bound (or indeed upper bound) of the left identity...
A: As a complement of @Calvin Khor's answer, I have verified the lower bound of $\left\|\sum_{k=1}^N \frac1{|x-k|}\right\|^*_{L^1_w}$.
For convenience consider $\left\|\sum_{k=1}^{2N} \frac1{|x-k|}\right\|^*_{L^1_w}$. For every $i \le N$ and $x \in [i-\frac{1}{2},i+\frac{1}{2}]$, we have $\frac{1}{\vert x-k\vert} > \frac{1}{k-i+1}$ for $k>i$. So $\left\vert \sum_{k=i+1}^{i+N} \frac1{|x-k|}\right\vert > \sum_{k=i+1}^{i+N}\frac{1}{k-i} \approx \ln(N)$. Therefore $\left\|\sum_{k=1}^{2N} \frac1{|x-k|}\right\|^*_{L^1_w} = \sup(t\lambda_f(t)) \ge \ln(N)\lambda_f(\ln(N))\ge N\ln(N)$. The upper bound may be similar but more complicated.
