- #1

joypav

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**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.**