How can we compare the lower envelope of a function with the function itself?

[joypav]

Problem:
Let f be a real-valued function defined on [a,b]. We define the lower envelope of f to be the function g defined by
$g(y) = sup_{\delta>0}inf_{\left| x-y \right|<\delta} f(x)$

(a) Show that for each $x \in [a,b]$, $g(x) \leq f(x)$ and $g(x) = f(x)$ if and only if f is lower semi-continuous at x.

(b) If f is bounded, then g is lower semi-continuous.

(c) If $\phi$ is any lower semi-continuous function such that $\phi(x) \leq f(x)$ for all $x \in [a,b]$, then $\phi(x) \leq g(x)$ for all $x \in [a,b]$.

Proof:
Part (a).
I am confused by the definition for the lower envelope. How do we compare g(x) and f(x)? I am confused about what x would be approaching in our definition for g(y), which is basically the liminf.

Part (b).
Let $\epsilon > 0$.
Then, $\exists \delta > 0$ such that,
$f(x) > g(x_0) + \epsilon$ for $x \in (x_0-\delta, x_0+\delta)$

Let $x \in (x_0-\delta, x_0+\delta)$ be given and let $n = \delta - \left| x-x_0 \right|$.
Then,
$g(x) = lim_{y \rightarrow x}inf f(y) \geq inf_{y \in (x_0-\delta, x_0+\delta)} f(y) \geq g(x_0) + \epsilon$

x arbitrary in $ (x_0-\delta, x_0+\delta) $
$\implies lim_{x\rightarrow x_0}inf g(x) \geq inf_{x \in (x_0-\delta, x_0+\delta)} g(x) \geq g(x_0) + \epsilon$

$\epsilon$ arbitrary
$\implies lim_{x\rightarrow x_0} g(x) \geq g(x_0)$.

 

[GJA]

Hi joypav,

I agree that it can be a bit confusing on how to compare $g(x)$ with $f(x)$, especially since we are using $y$ to define the function $g$ and $x$ as a dummy variable in $g$'s definition. Nevertheless, in order to match the notation of part (a), let's first make the changes $y \mapsto x$ and $x\mapsto t.$ Now $t$ is the dummy variable instead of $x$ so that $$g(x)=\sup_{\delta >0}\inf_{|t-x|<\delta}f(t).$$ Now, note that for each $\delta >0$, $$\inf_{|t-x|<\delta}f(t)\leq f(x), \qquad\qquad (*)$$ because $t=x$ will satisfy $|t-x|<\delta$ for all $\delta >0.$ Thus, since $(*)$ holds for all $\delta >0,$ $f(x)$ is an upper bound for the set $$\left\{\inf_{|t-x|<\delta} f(t):\delta >0\right\}.$$ Hence, $f(x)$ must be greater than or equal to the supremum over this set; i.e., $$g(x)\leq f(x).$$

 

[math771]

This shows that g is lower semi-continuous.

Part (c).
Let $x \in [a,b]$ be given.
Then,
$\phi(x) \leq f(x) \leq g(x)$.
Since $\phi$ is lower semi-continuous,
$lim_{y\rightarrow x} \phi(y) \leq \phi(x) \leq g(x)$.
Since $\phi(x) \leq g(x)$, we can take the limit as $y \rightarrow x$ and conclude that
$\phi(x) \leq g(x)$ for all $x \in [a,b]$.