4
$\begingroup$

Let $f: \mathbb{R}\rightarrow\mathbb{R}$ continuous and bounded below.

Show that there exist $x_0 \in \mathbb{R}$ such that $\forall x\ne x_0$, $$f(x_0)-f(x)<\vert x_0-x\vert$$

Since $f$ is continuous and bounded below then $m=\inf_{x\in R}f(x)$ exist and then :

$\forall \varepsilon>0, \exists x \in\mathbb{R}$ such that $m\le x<m+\varepsilon$

Finally I tried to use the continuity of $f$ but unfortunately the main problem is that I don't know where I am going.

$\endgroup$
1
$\begingroup$

Choose $c>1$, some $x_0$ and let $f_c(x) = f(x)+c|x-x_0|$.

Since $f$ is bounded below, we have $\lim_{|x| \to \infty} f_c(x) = \infty$, hence $f_c$ has a minimum at, say, $x'$ (this implicitly relies on compactness).

Then $f_c(x') \le f_c(x)$ for all $x$, or equivalently, $f(x')-f(x) \le c|x-x_0|-c|x'-x_0|$ for all $x$. The triangle inequality gives $|x-x_0| \le |x-x'| + |x'-x_0|$, hence we obtain $f(x')-f(x) \le c |x-x'|$ for all $x$.

Since $c>1$, we have $f(x')-f(x) < |x-x'|$ for all $x \neq x'$.

$\endgroup$
1
$\begingroup$

The $C$ be the area under (and including) the graph of $f(0)-|x|$ and let $G$ be the graph of $f$. Both these sets are closed, because these functions are continuous.

Therefore $C\cap G$ is closed, but it's also bounded, because $f$ was bounded from below. So it's compact and there's a point in $C\cap G$ with the lowest possible $y$-coordinate. Let it be $(x_0,f(x_0))$.

It's easy to see that the area under (and including) $f(x_0)-|x-x_0|$ is entirely in $C$ (draw it), so $f(x)>f(x_0)-|x-x_0|$ for all points $(x,f(x))$ outside of $C$. For the points inside, there's only one match, and that's $x=x_0$ by the choice of $x_0$.

$\endgroup$
  • $\begingroup$ @Nico I don't know. But is compactness really that advanced? You need a compactness argument to show e.g. that a continuous function on [0,1] has a minimum. $\endgroup$ – user2345215 Apr 9 '14 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy