11/07/2009, 09:31 AM
Let me rephrase in my words:
We consider the linear functions \( g_k \) on (k-1,k) determined by \( g_k(k-1)=f(k-1)={^{ k-1}b} \) and \( g_k(k)=f(k)={^k b} \), they are given by:
\( g_k(x) = ({^k b} - {^{ k-1}}b)(x-k) + {^k b} \).
As f is concave \( f(x+k) \ge g_k(x+k) \) for \( x\in (-1,0) \), and as \( \log_b \) is strictly increasing we have also \( f(x) \ge \log_b^{\circ k} g_k(x+k) \).
On the other hand we know that \( f(x+k)=\exp_b^{\circ k}(f(x))\uparrow a \) for \( k\to\infty \), hence \( a > f(x+k) > g_k(x+k) \) and \( g_k(x+k)\uparrow a \) for \( k\to\infty \).
We have now \( f(x+k)-g_k(x+k)\downarrow 0 \) for \( k\to\infty \). But I think that does not directly show the convergence \( f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0 \).
Any ideas?
We consider the linear functions \( g_k \) on (k-1,k) determined by \( g_k(k-1)=f(k-1)={^{ k-1}b} \) and \( g_k(k)=f(k)={^k b} \), they are given by:
\( g_k(x) = ({^k b} - {^{ k-1}}b)(x-k) + {^k b} \).
As f is concave \( f(x+k) \ge g_k(x+k) \) for \( x\in (-1,0) \), and as \( \log_b \) is strictly increasing we have also \( f(x) \ge \log_b^{\circ k} g_k(x+k) \).
On the other hand we know that \( f(x+k)=\exp_b^{\circ k}(f(x))\uparrow a \) for \( k\to\infty \), hence \( a > f(x+k) > g_k(x+k) \) and \( g_k(x+k)\uparrow a \) for \( k\to\infty \).
We have now \( f(x+k)-g_k(x+k)\downarrow 0 \) for \( k\to\infty \). But I think that does not directly show the convergence \( f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0 \).
Any ideas?
