# linear stability analysis of basic Solow model

I'm currently reading Romer's advanced macroeconomics, and encounter a question of linear stability of the model at the steady state.

We have the key equation $$k'(t)=sf(k(t))-(n+g+\delta)k(t)$$, where $$k$$ is capital per effective labour, $$n,g,\delta$$ each represents constant growth rate of labour, knowledge and depreciation rate of capital, respectively.

Use linear stability analysis, steady state is where $$k'(t)$$ is zero, denoted by $$\bar k$$. We can check the nature of the steady state by using $$sf'(\bar k)-(n+g+\delta)$$. We want to prove that the sign of it is negative.

We know $$f'(k) >0$$ and $$f''(k)<0$$, also other assumptions of $$f$$ given in the Solow model. Is it possible to prove it mathematically?

You outline the assumption that $$f$$ is strictly concave in $$k$$, so the average slope $$\frac{f(k) - f(0)}{k - 0}$$ is larger than $$f'(k)$$.
I will also use the frequently made assumption that $$f(0) = 0$$, though $$f(0) \geq 0$$ would be sufficient.
Taking these two together, we get $$f'(k) < \frac{f(k) - f(0)}{k - 0} = \frac{f(k)}{k}$$ or $$f'(k)k < f(k).$$ In the steady state $$sf( \bar k ) - (n+g+\delta) \bar k = 0.$$ Substituting in our inequality $$sf'( \bar k ) \bar k - (n+g+\delta) \bar k < 0.$$ if $$\bar k \neq 0$$, so we are not in the 'trivial' steady state, we can divide by $$\bar k$$ and we get $$sf'( \bar k ) - (n+g+\delta) < 0.$$
In the case of $$f(0) < 0$$ you would have a steady state where $$f(k)$$ intersects the line $$(n+g+\delta)k$$ from below, with a higher slope. This steady state would indeed be unstable. The same is true for $$\bar k = 0$$ when $$f(0) = 0$$.