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$$g_n(t)=\int_0^1\cos^3 (t+nu)h(u)\;du$$

I noticed that $$\left|g_n(t)-g_n(t_0)\right|\le\sup_{u\in[0,1]}|h(u)|\int_0^1\left|\cos^3(t+nu)-\cos^3(t_0+nu)\right|\;du$$

But how do I prove that this one can be arbitrarily small when $t\to t_0$?

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Nice question :) –  mick Oct 3 '12 at 13:16
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2 Answers 2

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As $a^3-b^3=(a-b)(a^2+ab+b^2)$, we have $$|g_n(t)-g_n(t_0)|\leq 3\lVert h\rVert_{\infty}\int_0^1|\cos(t+nu)-\cos(t_0+nu)|du.$$ As $\cos$ is a $1$-Lipschitz continuous map, $$|g_n(t)-g_n(t_0)|\leq 3\lVert h\rVert_{\infty}\|t-t_0|,$$ and we are done.

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Estimate the integrand using the mean value theorem.

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