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How can we prove -

$$|\dot{u}(x) - \tilde{u_h}(x)| \leqslant h \left[\max_{0 \leqslant y\leqslant1}\ddot{u}(y)\right],$$

where $0\leqslant x\leqslant 1$ and $\tilde{u_h}$ is interpolant of $u$.

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Hello Vineet Maheshwari. Could you precise what you mean by "interpolant of $u$"? – Sebastien B Jan 27 '13 at 18:06
I guess there is an error in your statement, you shall have a derivative too for the interpolant, so if you don't fix it, it is difficult to help you. – Sebastien B Jan 27 '13 at 18:34

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