0
$\begingroup$

Let $\phi:\mathbb{R}\to\mathbb{R}$ a continuous function. Fix $x_0\in\mathbb{R}$ and consider $$\psi:\mathbb{R}\to\mathbb{R},\ \psi(y)=\min_{\xi\in[x_0,y]}\phi(\xi)\ .$$ Is $\psi$ a continuous function? In particular does $\psi(y)\to\phi(x_0)$ as $y\to x_0$?

I think in general these questions has a negative answer: think to the function $$\phi(x)=\sin\frac{1}{x},\ \text{if } x\neq0 \quad,\quad \phi(0)=0\ $$ at point $x_0=0$.

But if we assume that $\phi$ has bounded variation on compact intervals, or that $\phi$ is $C^1$, can we hope to obtain a positive answer?

Edit after richard's comment The counter-example is not valid since the function is not continuous at $0$. So maybe is the hypothesis of continuity of $\phi$ sufficient?

Improve this question
$\endgroup$
3
  • $\begingroup$ At the beginning you assumed that $\phi$ is continuous everywhere, but in your example $\phi$ is discontinuous at $0$, why? $\endgroup$
    – qianzhang
    Apr 12, 2013 at 19:18
  • $\begingroup$ You are right, the counter-example is not valid. So maybe there is no need of further assumption? $\endgroup$
    – user22980
    Apr 12, 2013 at 19:25
  • 4
    $\begingroup$ $\forall \epsilon > 0$, $\exists \delta > 0$, $\forall x \in (x_0-\delta, x_0+\delta)$: $\phi(x) \in (\phi(x_0)-\epsilon, \phi(x_0)+\epsilon)$. So $\psi(y) \in (\phi(x_0)-\epsilon, \phi(x_0)+\epsilon)$ too, therefore $\psi(y) \to \phi(x_0)$. The same is true for all other points. $\endgroup$
    – user26107
    Apr 12, 2013 at 20:43

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.