Let $$X^k (t) := X^0 (t+t_k) - X^0 (t_k)$$ where $X^0(t)$ is the piecewise linear interpolation of $X^0(t_k) \equiv X_k=\sum_{i=0}^{k-1} a_i b_i$ with interpolation intervals $a_k$. $t_k=\sum_{i=0}^{k-1} a_i$, $t_0=0$.

So we can think of supporting points $t_i$ and distances $a_k=t_{k+1}-t_k$.

I know $b_k \rightarrow 0$, $k \rightarrow \infty$. And I want to show that $X^k(t) \rightarrow 0$ uniformly on bounded intervals. As conditions on $a_k$ I have $\sum_{i=0}^{\infty} a_i=\infty, a_i \rightarrow 0, a_i>0$.

So when I start to look at it, I see $X^0(t_k) \equiv X_k \rightarrow 0$ and the piecewise linear interpolation between these $X_k$ that tend to 0 should also tend to 0. But how can I do this more formally and with uniform convergence? Maybe it is only about finding an expression for $X^0 (t+t_k)$ independent to which interpolation interval $t+t_k$ belong, but I don't really see how this could be done.

Note: $X^k$ is defined on $R$ and should converge to $0$ on $R$, and if we restrict $X^k(t)$ to a bounded interval $I$, then one wants uniform convergence.

  • $\begingroup$ @Theo: The expression "finite intervals" comes from the literature I use (Kushner and Clark 1978, Stochastic Approximation Methods for Constrained and Unconstrained Systems, Proof of 2.3.1) $\endgroup$
    – Johannes L
    Sep 7, 2011 at 7:19
  • $\begingroup$ Thanks for the clarification. I still think that "bounded" is much more common (at least in the literature I use). $\endgroup$
    – t.b.
    Sep 7, 2011 at 10:36


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