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I have a piecewise linear interpolation:

$$ B(t) = \frac{t_{l+1}-t}{t_{l+1}-{t_{l}}} B_l + \frac{t-t_l}{t_{l+1}-{t_{l}}} B_{l+1} \quad \text{ if $t \in (t_l, t_{l+1})$;}$$ $B(t_l)=B_{l}$ and $B(t) = B_0$ for $t \leq0$.

I want to use that $B(t)$ is uniformly continuous on whole $R$. Is there an easy argument for that? (For example if from continuity on R and uniform continuity everywhere but in $t_1, t_2, \cdot$ (countable many points) would follow uniform continuity everywhere). Or maybe there is an elegant way via Lipschitz-continuity? I think the uniform continuity inside the intervals $(t_l, t_{l+1})$ is clear, it's just the question how to put it all together without having to recede to the $\epsilon$-$\delta$-definition.

Q: Is there a better (shorter/more elegant) way to follow uniform continuity here? example

The red line would be a line with slope = supremum of the slopes on the distinct intervals, the orange line is a sample secant line.

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So I have to add the condition $B_l \rightarrow 0$, $l \rightarrow \infty$, otherwise the uniform continuity should even have counterexamples – Johannes L Sep 6 '11 at 10:01

Your approach via Lipschitz continuiuty may work: The piecewise linear interpolation is Lipschitz continuous with Lipschitz constant equal to the largest slope within the interpolation intervals - provided this exists. Hence, your function is Lipschitz continuous if the slopes on the interpolation intervals are uniformly bounded. Otherwise, your function will not even be uniformly continuous.

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I guess with $L= \sup_l (\text{slope in }(t_l, t_{l+1}))$ it should work. This doesn't have to hold, but in my application I know $B_l \rightarrow 0$ as well, and then it should fine. – Johannes L Sep 6 '11 at 9:57

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