Consider a simplified version of the Solow-Swan model in discrete time, where technology is normalized to one and the population size is constant.

The Solow equation is given by \begin{equation} k_{t+1} = s f(k_t) + (1-\delta) k_t \end{equation} where $k_t \equiv \frac{K_t}{L}$ represents the capital per worker, $s \in (0,1)$ is the saving rate, $f(k)$ is the production function in intensive form, and $\delta \in (0,1)$ is the depreciation rate of physical capital.

I attempt to prove global stability as follows:

First, a steady state $k^{\ast}$ is a fixed point of the law of motion of capital: $k^{\ast} = g(k^{\ast})$, and it is about: \begin{equation} \frac{k^{\ast}}{f(k^{\ast})} = \frac{s}{\delta} \end{equation}

The existence of a steady state $k^{\ast}$ satisfying the equation above is guaranteed by the intermediate value theorem since $g(k)$ is continuous over its domain and defined over a closed interval, say $[a,b]$, with $g(a) \ne g(b)$. The uniqueness (discarding the trivial steady state, $k^{\ast}=0$) stems from the fact that $g(k)$ is strictly increasing (yet bounded above due to the Inada conditions, i.e., $f'(\infty) = 0$).

Now, I prove global stability as follows:

The function $g(k) = s f(k) + (1-\delta) k_t$ satisfies: $i) g'(k) > 0$ and $ii) g''(k) < 0$. Consequently, $g(k) > k$ if and only if $k < k^{\ast}$. Similarly, $g(k) < k$ if and only if $k > k^{\ast}$. The global stability is due to the diminishing marginal return of capital. To see this, let $\gamma$ denote the growth of $k$:

\begin{equation} \gamma = \frac{k_{t+1} -k_t}{k_t} = s \frac{f(k_t)}{k_t} - \delta = \frac{s}{z(k_{t})} - \delta \end{equation} Here, $z(k_{t}) = \frac{k_t}{f(k_t)}$ is strictly increasing in $k_t$, and $z(k^{\ast}) = s/\delta$. Trivially, the growth rate of capital per worker at $k^{\ast}$ is zero. It is evident that if $k' > k^{\ast}$, negative growth in $k$ occurs until capital per worker converges to $k^{\ast}$. Similarly, if $k' < k^{\ast}$, positive growth in $k$ occurs until it converges to $k^{\ast}$.

Is the proof presented in the following manner sufficiently rigorous, or do you believe there is room for improvement?

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    $\begingroup$ What I was trying to say is that the existence of a fixed point is guaranteed by the intermediate value theorem. The uniqueness of $k^{\ast}$ is guaranteed by the fact that $g(k)$ is strictly increasing, and the global stability comes from the diminishing returns of capital. Are these statements correct? $\endgroup$
    – Maximilian
    Mar 2 at 9:49
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    $\begingroup$ Hi Maximilian! Thank you for the clarification. Please do not edit your question in substantive ways, SE is a poor fit for this sort of iterative processes, because the answers become out of sync with the questions. On your new proof: the sign of the change of $k$ does not prove convergence. For an example see Xeno's/Achilles and the tortoise paradox. The good news is that you do not have to prove convergence, that is asymptotic stability. You can use the signs to argue that the distance does not increase, and that should be fine for global stability. $\endgroup$
    – Giskard
    Mar 2 at 21:12
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    $\begingroup$ @Maximilian Just a little observation. The use of the intermediate value theorem is not completely correct, that $g(a)\neq g(b)$ is not sufficient, you sholud show that there are $a$ and $b$ such that $g(a)<\delta/s$ and $g(b)>\delta/s$. It is better to use the version of the theorem with inf and sup and use the Inada conditions. $\endgroup$ Mar 2 at 21:38
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    $\begingroup$ @Maximilian As for the proof of stability, your argument is correct and well done, but it is not a sufficient rigorous mathematical proof. To prove stability we need to prove that every sequence $k (t)$ converges to $k^*$. But to prove, for example, that if $k(t)<k^*$, $k_{t+1}>k(t)$ doesn't ensure stability: $k(t+1)$ can go far beyond $k^*$ and then jump again on the left of $k^*$, and the sequence can oscillate around $k^*$, going further and further away from $k^*$. For a rigorous proof see Acemoglu, Modern Economic Growth. $\endgroup$ Mar 2 at 21:55
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    $\begingroup$ A classical reference for economic dynamics is Gandolfo, Economic dynamics, where you find also difference equations, that is discrete time equations. As for a matematical book about dynamics systems in general, they are often advanced, but there is a very interesting book, not advanced, no particular mathematics is required, but it is not addressed to economists, so there is a lot of subjects that are not very important for economics, the book is Strogatz, Non linear Dynamics and Chaos. $\endgroup$ Mar 3 at 11:33

2 Answers 2


The assertion that $k^{\ast}$ is the fixed point follows from the intermediate value theorem (as $g(k)$ is continuous over its domain and $g(k') \neq g(k'')$ for $k' \neq k''$). The uniqueness stems from the fact that $g(k)$ is strictly increasing

Your approach about the proof of existence and uniqueness is substantially correct, but it needs some specifications, in particular about the function to which the intermediate value theorem is applied and the use of the Inada conditions.

Existence of a fixed point

It must be specified to which function we apply the intermediate value theorem.

The existence of a steady state equilibrium means that there exists a constant $k^*$ such that $$k(t+1)= g(k(t))= s f(k(t)) +(1- \delta) k(t) = k(t)\tag {1}.$$

Therefore, a steady state value $k^*$ must satisfy, for $k\neq 0$$^1$:

$$\frac {f(k)}{k}= \frac {\delta}{s} \tag {2}$$

We can apply the intermediate value theorem to the function $\frac {f(k)}{k}$ above.

Here, the Inada conditions are crucial. In fact, applying the Inada conditions and the De l'Hôpital rule, we have:

$$\lim_{k\rightarrow 0} {\frac {f(k)}{k}}= \lim_{k\rightarrow 0}{\frac {f'(k)}{1}}= +\infty \tag {3}$$

$$\lim_{k\rightarrow +\infty} {\frac {f(k)}{k}}= \lim_{k\rightarrow 0}{\frac {f'(k)}{1}}= 0,\tag {4}$$

where the second equalities of equations $(3)$ and $(4)$ are the Inada conditions.

The intermediate value theorem says that a continuous function on an interval $I$ takes all the values between its $\inf$ and its $\sup$.$^2$

Here, $\inf {\frac {f(k)}{k}}=0$ and $\sup {\frac {f(k)}{k}}=+\infty$, so that $\frac {f(k)}{k}$ takes all values in $\mathbb{R}^+$, including $\frac {\delta}{s}$

Therefore, a steady state equilibrium exists.


However, I can’t understand why you said:

and $g(k') \neq g(k'')$ for $k' \neq k''$.

This means just that the function $g$ is injective, what we already know as it is strictly increasing.

But what has this to do with existence and intermediate value theorem? The intermediate value theorem doesn’t require the function to be injective, and the proof above will hold even if $g$ wouldn't be injective.

Consider the usual graph, in which we have a not injective $g$: this evidently does not affect the existence of a steady state, but only its unicity:

enter image description here $$\;$$


Your statement concerning uniqueness is correct, we in particular observe that, because of the strict monotonicity of $g(k)$ (and $g''<0$), for $k<k^* $ $g(k)>k$, and for $k>k^*$ $g(k)>k$. Therefore, the only point in which $g(k)=k$ is $k^*$.


A qualitative argument to establish stability is to consider that if $k(t)$ is above $k^*$, it decreases, and if it is below $k^*$ it increases, and this is related to diminishing returns of capital.

A formal, analitycal proof of stability, however, requires theorems concerning the stability of discrete dynamical systems, much more complicated.

$^1$ Of course, there is a trivial fixed point, for $k=0$, but we discard it as of no economic interest, and moreover unstable.

$^2$ I specify this version of the intermediate value theorem, with $\inf$ and $\sup$, because often we find a different version, where the values of the function at the extreme of a closed interval $[a,b]$ are specified, but the two versions are equivalent.

  • $\begingroup$ Hi! You seem to be proving that the equilibrium exists and is unique. Where do we see global stability? $\endgroup$
    – Giskard
    Mar 2 at 18:14
  • $\begingroup$ @Giskard, if you read the question, it is about existence and uniqueness, stability is in the title only. There is no question in the text about stability. Maybe the OP asks for proofs of existence and uniqueness as a prerequisite for studyng stability? I don't know. $\endgroup$ Mar 2 at 18:18
  • $\begingroup$ Hmm, what is this then? "I attempt to prove global stability as follows:" $\endgroup$
    – Giskard
    Mar 2 at 18:20
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    $\begingroup$ @giskard I brought back the comments, please continue the discussion in the chat $\endgroup$
    – 1muflon1
    Mar 2 at 19:56
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    $\begingroup$ I made the question more clear. Sorry for the confusion $\endgroup$
    – Maximilian
    Mar 2 at 20:33

A unique fixed point may not be stable. For an example, check the dynamic system $$ x_{t+1} = 2x_t, $$ where the unique fixed point is $x^* = 0$, but this is not stable, in fact, starting from any point other than 0 you will get a trajectory that diverges from the fixed point. To prove global stability, you need to show that no trajectories diverge from the fixed point, show that their distance is not increasing.


The same equations can be used, but this is what you need to prove and should be working towards.


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