Problem with convergence of ReLu in MLP
I will start with a toy example for the convergence part. Suppose that the loss function is $f(x) = x^4$ and we want to minimize it using the gradient descent. Clearly, the minimum is attained at zero and, in general, we would like the magnitude of the current approximation to decrease.
The update rule of gradient descent is
$$ x_{k+1} = x_k - \lambda \nabla f = x_k - \lambda \cdot 4x_k^3.$$
Simplifying the expression, we get
$$x^{k+1} = x^k(1 - 4\lambda x_k^2).$$
And now the combo initialization + learning rate start to appear. If $|1-4\lambda x_0^2| &lt; 1$, then $|x_0| &gt; |x_1| &gt; |x_2| &gt; ...$. The sequence will go to zero eventually. If $|1-4\lambda x_0^2| &gt; 1$, then $|x_0| &lt; |x_1|$. In this case, $|x_1| &lt; |x_2|$ and so on -- the sequence will grow. Therefore, if the learning rate $\lambda$ if fixed, then the initial values $x^0$ determine if the grad descent converges or not.
When the gradient descent converge?
The math says that the gradient descent converge when the learning rate $\lambda$ and the gradient $\nabla f$ satisfy
$$\|\lambda \nabla f\| &lt; 1$$
along the optimization path. Note that this condition does not need to be valid for all values of $x$, but at every $x_0, x_1, x_2, ...$
For many &quot;good&quot; functions, it suffices to require $\|\lambda \nabla f\| &lt; 1$ only at $x_0$. The reason is that after the first iteration, we are closer to the local minimum, and for many &quot;good&quot; functions, it means that the gradient will be smaller.
What about weight choice in [-0.5,0.5] and [-1, 1]?
I think of it as follows: Suppose that we selected weights in $[-0.5, 0.5]$ (model 1) and then multiplied all weights by 2 to get uniform distribution in $[-1, 1]$ (model 2). Suppose that the learning rate is identical in both cases, and let's check how the SGD performs. For simplicity of the argument, I replace it with the gradient descent.
How does it transfer to the NN?
Note that a linear map (say, for the dense layer) has the following property. Suppose that all weights W are multiplied by 2, then
$$\|2W\| = 2\|W\|.$$
ReLU is quazi-linearly-scalable: for every $a &gt; 0$,
$$a\cdot\text{ReLU}(x)=\text{ReLu}(ax).$$
Note that if your NN has $d$ layers, then, multiplying all weights by 2, your output increases in $2^d$ times (factor of 2 for each layer).
It is hard to compute the gradient for the NN, but using the product rule, I expect it to increase in approximately $2^{d}$ times. If model 1 satisfies the conditions for the gradient descent convergence, then we need to decrease the learning rate about $2^{d}$ times to guarantee the gradient descent convergence condition.