Non-trivial steady state

Let's guess that the value function is of the form $a + b ln(k)$.

Then substituting for $V(k) = a + b ln(k)$ in the Bellman equation gives: $$ a + b ln(k) = max_{k'}left(ln(k^alpha - k') + beta(a + b ln(k')right) $$ The first order condition is given by: $$ begin{align*} &frac{-1}{k^alpha - k'} + beta b frac{1}{k'} = 0, to & k' = beta b (k^alpha - k'), to & k' = frac{beta b}{1+ beta b} k^alpha end{align*} $$ If we plug this into the objective function of the Bellman equation, we obtain the following identity: $$ begin{align*} a + b ln(k) &= lnleft(k^alpha - frac{beta b}{1 + beta b}k^alpharight) + betaleft(a + b lnleft(frac{beta b}{1 + beta b}k^alpharight)right), &= (alpha + beta b alpha) ln(k) + lnleft(1 - frac{beta b}{1 + beta b}right) + beta a + beta b lnleft(frac{beta b}{1 + beta b}right) end{align*} $$ As this holds for all $k (> 0)$ we can equate coefficients on both sides: $$ begin{align*} a &= lnleft(frac{1}{1 + beta b}right) + beta a + beta b lnleft(frac{beta b}{1 + beta b}right), b & = alpha + beta b alpha end{align*} $$ The second one gives a closed form expression for $b$: $$ b = frac{alpha}{1 - beta alpha}. $$ Then substituting this into the first order condition gives: $$ begin{align*} k_{t+1} &= frac{beta frac{alpha}{1 - beta alpha}}{1 + beta frac{alpha}{1 - beta alpha}}k_t^alpha, &= beta alpha k^alpha_t tag{1} end{align*} $$

This shows that: $$ k_{t + 1} > k_t iff beta alpha k_t^alpha > k_t iff k_t < (beta alpha)^{frac{1}{1 - alpha}} $$ So the capital stock will rise as long as $k_t$ is below $(beta alpha)^{frac{1}{1 - alpha}}$ and it will decrease if $k_t$ it is above this threshold.

According to the dynamic equation (1) above, it would appear that $k = 0$ is also a steady state. However, for $k = 0$, the first order conditions are not satisfied and in fact the value function does not exist. Anyway, for $k_t$ very close, it's value will be below $(beta alpha)^{frac{1}{1 - alpha}}$ so the stock of capital should increase to the unique steady state.