Why are activation functions of rectified linear units (ReLU) considered non-linear?

$$ f(x) = \max(0,x)$$

They are linear when the input is positive and from my understanding to unlock the representative power of deep networks non-linear activations are a must, otherwise the whole network could be represented by a single layer.


1 Answer 1


RELUs are nonlinearities. To help your intuition, consider a very simple network with 1 input unit $x$, 2 hidden units $y_i$, and 1 output unit $z$. With this simple network we could implement an absolute value function,

$$z = \max(0, x) + \max(0, -x),$$

or something that looks similar to the commonly used sigmoid function,

$$z = \max(0, x + 1) - \max(0, x - 1).$$

By combining these into larger networks/using more hidden units, we can approximate arbitrary functions.

$\hskip2in$RELU network function

  • $\begingroup$ Would these types of hand-constructed ReLus be built apriori and hard coded in as layers? If so, how would you know that your network required one of these specially built ReLus in particular? $\endgroup$ Sep 16, 2016 at 7:53
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    $\begingroup$ @MonicaHeddneck You could specify your own non-linearities, yes. What makes one activation function better than another is a constant research topic. For example, we used to use sigmoids, $\sigma(x) = \frac{1}{1 + e^{-x}}$, but then due to the vanishing gradient problem, ReLUs became more popular. So it's up to you to use different non-linearity activation functions. $\endgroup$ Sep 19, 2016 at 21:02
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    $\begingroup$ How would you approximate $e^x$ with ReLU in out of sample? $\endgroup$
    – Aksakal
    Sep 12, 2018 at 21:42
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    $\begingroup$ @Lucas, So basically if combine(+) >1 ReLUs we can approximate any function, but if we simply reLu(reLu(....)) it will be linear always? Also, here you change x to x+1, that could be thought as Z=Wx+b where W & b changes to give different variants of such kind x & x+1? $\endgroup$
    – Anu
    Mar 31, 2019 at 0:12

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