I was hoping to understand what the smooth $l_1$ loss does, but I'm not able to find any good explanation of online, I know $l_1$ loss calculates the absolute error, but what is the use of smooth $L_1$, any answers would be helpful.

  • $\begingroup$ It tries to mimic the l1-loss (look at the graph), while being smooth. The smoothness property allows for treatment as smooth continuous optimization, which is in general easier than non-smooth opt. $\endgroup$
    – sascha
    Commented Jun 17, 2018 at 12:07

1 Answer 1


Smooth L1-loss can be interpreted as a combination of L1-loss and L2-loss. It behaves as L1-loss when the absolute value of the argument is high, and it behaves like L2-loss when the absolute value of the argument is close to zero. The equation is:

$L_{1;smooth} = \begin{cases}|x| & \text{if $|x|>\alpha$;} \\ \frac{1}{|\alpha|}x^2 & \text{if $|x| \leq \alpha$}\end{cases}$

$\alpha$ is a hyper-parameter here and is usually taken as 1. $\frac{1}{\alpha}$ appears near $x^2$ term to make it continuous.

Smooth L1-loss combines the advantages of L1-loss (steady gradients for large values of $x$) and L2-loss (less oscillations during updates when $x$ is small).

Another form of smooth L1-loss is Huber loss. They achieve the same thing. Taken from Wikipedia, Huber loss is

$ L_\delta (a) = \begin{cases} \frac{1}{2}{a^2} & \text{for } |a| \le \delta, \\ \delta (|a| - \frac{1}{2}\delta), & \text{otherwise.} \end{cases} $

  • $\begingroup$ Should be $|x|-0.5$ $\endgroup$
    – Tim
    Commented Sep 30, 2018 at 12:25
  • $\begingroup$ @Tim, can you please be more specific about the correction? $\endgroup$ Commented Oct 1, 2018 at 8:00
  • $\begingroup$ Try plotting it, the appropriate form is en.wikipedia.org/wiki/Huber_loss $\endgroup$
    – Tim
    Commented Oct 1, 2018 at 8:16
  • 1
    $\begingroup$ Oh yes, it needs to be continuous. I will include two varieties then $\endgroup$ Commented Oct 1, 2018 at 10:11

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