# Lower SemiContinuity Problem: Proofs & Step Funcs

• MHB
• joypav
In summary, a function $\phi$ defined on an interval $[a, b]$ is a step function if it only takes on one value in each subinterval of $[a, b]$. To show that a step function is lower semi-continuous, it must be proven that $\phi(x_i)$ is less than or equal to the smaller of the two values assumed in the subintervals adjacent to $x_i$. Conversely, if $\phi(x_i)$ is less than or equal to the smaller of the two values assumed in the adjacent subintervals, then the function is lower semi-continuous. This can be proven by considering two cases: when $x_0$ is within an interval and when $x_0 joypav Problem: (a) Show that$f: E \rightarrow R$is lower semi-continuous at$x_0 \in E$if and only if$f(x_0) \leq \liminf_{x \rightarrow x_0}f(x)$Proof: "$\rightarrow$" Assume f is l.s.c. at$x_0 \in E$.$\implies \forall \epsilon > 0, \exists \delta >0$such that if$\left|x-x_0 \right|<\delta$, then$f(x_0)-\epsilon \leq f(x)$Consider$\liminf_{x \rightarrow x_0}f(x) = \liminf_{n \rightarrow \infty}\left\{f(x): \left| x-x_0 \right|<1/n\right\}$. Let$y \in R, y = \liminf_{x \rightarrow x_0}f(x) \in R$. Then$y = f(x')$for some$x' \in E$. Because f is l.s.c. at$x_0$,$\forall \epsilon > 0, \exists \delta >0$such that if$\left|x'-x_0 \right|<\delta$, then$f(x') \geq f(x_0)- \epsilon\implies \liminf_{x \rightarrow x_0}f(x) \geq f(x_0) - \epsilon$Let$\epsilon \rightarrow 0\implies \liminf_{x \rightarrow x_0}f(x) \geq f(x_0)$. "$\leftarrow$" Assume$f(x_0) \leq \liminf_{x \rightarrow x_0}f(x)$. I am thinking I should proceed with proving by contradiction? Meaning, assume there is an$x_0$where f is not l.s.c. (b) A real-valued function$\phi$defined on an interval$[a, b]$is called a step function if there is a partition$a = x_0 < x_1 <...< x_n = b$such that for each i the function$\phi$assumes only one value in the interval$(x_{i−1}, x_i)$. Show that a step function$\phi$is lower semi-continuous if and only if$\phi(x_i)$is less than or equal to the smaller of the two values assumed in$(x_{i−1}, x_i)$and$(x_i, x_{i+1})$. For this part I think I am just confused by the definition given for step function. However, I am still thinking about the problem. So I will add onto the post if I make some headway. Please state the definition you're using for l.s.c. functions. Also, are we assuming that$E$is a subset of$\Bbb R$? Euge said: Please state the definition you're using for l.s.c. functions. Also, are we assuming that$E$is a subset of$\Bbb R$? Sorry! Yes, E is a subset of R. The definition we were given is what I used in the proof: A function f is l.s.c. on E if$\forall x_0 \in E$,$\forall \epsilon > 0, \exists \delta >0$such that if$\left|x-x_0 \right|<\delta$, then$f(x_0)-\epsilon \leq f(x)$You do not seem to have the correct formulation of the inferior limit$\liminf\limits_{x\to x_0} f(x)$. If$x_0$is a limit point of$E$,$\liminf\limits_{x\to x_0} f(x)$is defined as $$\sup\limits_{\delta > 0}\, \inf\limits_{x\in E,\, 0 < |x - x_0|< \delta} f(x)$$ which is the same as $$\lim\limits_{\delta \to 0}\, \inf\limits_{x\in E,\, 0 < |x - x_0| < \delta} f(x)$$ Now suppose$f$is l.s.c. at$x_0$. If$\epsilon > 0$, there exists a$\delta > 0$such that for all$x$,$|x - x_0| < \delta$implies$f(x_0) - \epsilon \le f(x)$. So for$0 < \eta < \delta$,$f(x_0) - \epsilon$is a lower bound for the of values$f(x)$where$x\in E$and$0 <|x - x_0| < \eta$. Hence$f(x_0) - \epsilon \le \inf\{f(x):x\in E,\, 0 < |x - x_0| < \eta\}$for all$0 < \eta < \delta$. Therefore$f(x_0) - \epsilon \le \liminf\limits_{x\to x_0} f(x)$. As$\epsilon$was arbitrary,$f(x_0) \le \liminf\limits_{x\to x_0} f(x)$. Conversely, suppose$f(x_0) \le \liminf\limits_{x\to x_0} f(x)$. Set$L = \liminf\limits_{x\to x_0} f(x)$. Fix$\epsilon > 0$, and choose a positive number$\delta$such that for all$x\in E$,$0 < |x - x_0| < \delta$implies$f(x) > L - \epsilon$. By assumption$f(x_0) \ge L$, so$f(x) > f(x_0) - \epsilon$whenever$|x - x_0| < \delta$. Hence,$f$is l.s.c. at$x_0$. Thank you! Here is what I have for part b. "$\leftarrow$" Assume$\phi(x_i)$is less than or equal to the smaller of the two values assumed in$(x_{i-1},x_i),(x_i,x_{i+1})$. Case I:$x_0 \in (x_{i-1},x_i)$, some i Then, choose$\delta = \frac{x_i-x_0}{2}$If$\left| x-x_0 \right| < \delta = \frac{x_i-x_0}{2} \implies x \in (x_{i-1},x_i) \implies \phi(x)=\phi(x_0)$Then,$\forall \epsilon > 0, \phi(x_0) - \epsilon \leq \phi(x_0) \implies \forall \epsilon > 0$and for$\delta = \frac{x_i-x_0}{2}, \phi(x_0) - \epsilon \leq \phi(x)\implies \phi$is lower semi-continuousCase II:$x_0$is an endpoint of an interval, WLOG say$x_0 = x_i$, some i for$\forall x \in (x_{i-1}, x_i), \phi(x) = a$for$\forall x \in (x_i, x_{i+1}), \phi(x) = b$WLOG, assume a<b. Then, by assumption,$\phi(x_i) \leq a$. Then, for$\delta = x_i - x_{i-1}$,$\forall \epsilon>0$, if$\left| x-x_0 \right| = \left| x-x_i \right| < \delta \implies x \in (x_{i-1}, x_i)$or$x \in (x_i, x_{i+1})$if$x \in (x_{i-1}, x_i)$, then$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon \leq a - \epsilon < a = \phi(x) \implies \phi(x_0) - \epsilon \leq \phi(x)$if$x \in (x_i, x_{i+1})$, then$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon \leq a - \epsilon < a < b = \phi(x) \implies \phi(x_0) - \epsilon \leq \phi(x)\implies \phi$is lower semi-continuous"$\rightarrow$" Assume$\phi$is lower semi-continuous$\implies$for every$x_0 \in [a, b], \forall \epsilon>0, \exists \delta>0, \left| x-x_0 \right|<\delta \implies \phi(x_0) - \epsilon \leq \phi(x)$By way of contradiction, assume$\phi(x_i)$is greater than the smaller of the two values assumed in$(x_{i-1},x_i),(x_i,x_{i+1})$. for$\forall x \in (x_{i-1}, x_i), \phi(x) = a$for$\forall x \in (x_i, x_{i+1}), \phi(x) = b$WLOG, assume a<b. Then, by assumption,$\phi(x_i) > a$. Choose$x_0=x_i$. Then, if$x<x_0$,$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon > a - \epsilon = \phi(x) - \epsilon\implies \phi(x_0) - \epsilon > \phi(x)$, a contradiction$\implies \phi(x_i) \leq a\$

## What is the Lower SemiContinuity Problem?

The Lower SemiContinuity Problem is a mathematical concept in functional analysis that deals with the behavior of functions on a topological space. Specifically, it examines whether a function remains continuous when its domain is restricted to a subset of the topological space.

## What does it mean for a function to be lower semi-continuous?

A function is lower semi-continuous if, for any given point in its domain, all values of the function that are close to that point are also close to the value of the function at that point. In other words, the function does not have any abrupt changes or discontinuities.

## What is the importance of the Lower SemiContinuity Problem?

The Lower SemiContinuity Problem has applications in various fields such as optimization, economics, and physics. It helps in understanding the behavior and properties of functions, which is crucial in solving real-world problems and making accurate predictions.

## What are some common techniques used to prove lower semi-continuity?

Some common techniques used to prove lower semi-continuity include the use of sub-level sets, sequential continuity, and the liminf inequality. Other methods such as the use of epigraphs and convexity can also be employed depending on the specific problem at hand.

## What are step functions and how are they related to the Lower SemiContinuity Problem?

Step functions are piecewise constant functions that have a finite number of intervals where the function takes on different values. They are closely related to the Lower SemiContinuity Problem as they are often used as building blocks in constructing more complex functions that are lower semi-continuous.

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