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I wish to find the best way to prove that $\phi_n:= [\psi(x+{1\over n})-\psi(x)]n$ where $\psi$ is continuously differentiable on $(a,b)$, converges uniformly to $\psi'(x)$ on all closed subintervals of $(a,b)$.

It is clear that $(\phi_n)$ converges to $\psi'$. But not so clear that $(\phi_n')$ converges uniformly on every closed subinterval. I am thinking of using the continuity property of the derivative...? If I could show this then it follows that $\phi_n$ converges uniformly to $\psi'(x)$ on all closed subintervals of $(a,b)$.

Is there a better way to show this though?

Thanks.

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Hint: Use the mean value theorem on $\phi_n(x)$ for each $x$. –  Zarrax Sep 20 '11 at 1:42

1 Answer 1

We will follow Zarrax's suggestion to use the mean value theorem.

Fix a closed subinterval $X$ of $(a,b)$ and a parameter $\varepsilon > 0$. Since every continuous function on the reals is uniformly continuous on closed, bounded intervals, it follows that there exists $\delta > 0$ such that for all $x, y \in X$, $$ |y-x| \lt \delta \quad \implies \quad |\psi'(y) - \psi'(x)| \lt \varepsilon. \tag{$\ast$} $$ Finally, pick an integer $N > \frac{1}{\delta}$.

It remains to check that this choice of $N$ works. Fix any $n \geqslant N$. For any $x \in X$, by the mean value theorem on $\phi_n$, we have $$ \phi_n(x) = \frac {\psi_n \Big(x+\frac{1}{n}\Big) - \psi_n(x)}{ \frac{1}{n} } = \psi_n'(\xi_n) $$ for some $\xi_n \in \Big(x, x+\frac{1}{n} \Big)$. In particular, $|\xi_n - x| \lt \frac{1}{n} \leqslant \frac{1}{N} \leqslant \delta$. Therefore, from $(\ast)$, it follows that $|\psi'(\xi_n) - \psi'(x)| \lt \varepsilon$; in turn, this implies that $|\phi_n(x) - \psi'(x)| \lt \varepsilon$. That is, for all $n \geqslant N$, we have $\sup \{ |\phi_n(x) - \phi'(x)| \ :\ x \in X \} \leqslant \varepsilon$, and we are done.

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