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?

  • $\begingroup$ 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


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