# Lowe Semicontinuity (1 Viewer)

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Lower Semicontinuity

I found this in the web:
We say that $$f$$ is lower semi-continuous at $$x_0$$ if for every $$\epsilon > 0$$ there exists a neighborhood $$U$$ of $$x_0$$ such that $$f(x) > f(x_0) - \epsilon$$ for all $$x$$ in $$U$$. Equivalently, this can be expressed as

$$\liminf_{x \to x_0} f(x) \geq f(x_0).$$

The first definition is quite clear to me (by looking at an example of lower semicontinuity diagram). But I don't understand its equivalence to the second definition. Could someone draw the connection?

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#### HallsofIvy

I found this in the web:
We say that $$f$$ is lower semi-continuous at $$x_0$$ if for every $$\epsilon > 0$$ there exists a neighborhood $$U$$ of $$x_0$$ such that $$f(x) > f(x_0) - \epsilon$$ for all $$x$$ in $$U$$. Equivalently, this can be expressed as

$$\liminf_{x \to x_0} f(x) \geq f(x_0).$$

The first definition is quite clear to me (by looking at an example of lower semicontinuity diagram). But I don't understand its equivalence to the second definition. Could someone draw the connection?
Looks pretty straight forward to me. Suppose $$f(x) > f(x_0) - \epsilon$$ for all x in some neighborhood U. Let xn be a sequence converging to x0. Then eventually, it will be in U. Since we can ignore x's that are not in U, its limit must satisfy $$lim f(x_n)\geq f(x_0)$$ and so of course must lim inf.

Conversely suppose $$\liminf_{x \to x_0} f(x) \geq f(x_0)$$ and suppose there were no neighborhood U as above. Let Un be (x0- 1/n, x0+ 1/n). Since none of these can satisfy $$f(x) > f(x_0) - \epsilon$$ for all x in Un, there must exist xn in Un such that $$f(x) \leq f(x_0) - \epsilon$$. But then, for that sequence, $$lim f(x_n)\leq f(x_0)$$, contradicting $$\liminf_{x \to x_0} f(x) \geq f(x_0)$$.

Just in case some one out there is thinking "lower semi-continuous" must have something to do with "continuous", let me point out that the function f(x)= 1000 if x is not 0, 0 if x= 0 is lower semi-continuous at x=0!

HallsofIvy said:
Looks pretty straight forward to me. Suppose $$f(x) > f(x_0) - \epsilon$$ for all x in some neighborhood U. Let xn be a sequence converging to x0. Then eventually, it will be in U. Since we can ignore x's that are not in U, its limit must satisfy $$lim f(x_n)\geq f(x_0)$$ and so of course must lim inf.!
This is the part I don't understand. Suppose $$f(x) > f(x_0) - \epsilon$$ for all x in some neighborhood U. Then for some x in the neighborhood of U, $$f(x) < f(x_0)$$ may hold true since $$\epsilon >0$$. I am lost.

HallsofIvy said:
Conversely suppose $$\liminf_{x \to x_0} f(x) \geq f(x_0)$$ and suppose there were no neighborhood U as above. Let Un be (x0- 1/n, x0+ 1/n). Since none of these can satisfy $$f(x) > f(x_0) - \epsilon$$ for all x in Un, there must exist xn in Un such that $$f(x) \leq f(x_0) - \epsilon$$. But then, for that sequence, $$lim f(x_n)\leq f(x_0)$$, contradicting $$\liminf_{x \to x_0} f(x) \geq f(x_0)$$.
I agree with this one.

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Oh I see my problem now. The keyword that I missed was "for every $$\epsilon > 0$$, ...." .
I have one last question. How does one read $$\liminf_{x \to x_0}$$? Infimum of x at the limit point?

#### HallsofIvy

Oh I see my problem now. The keyword that I missed was "for every $$\epsilon > 0$$, ...." .
I have one last question. How does one read $$\liminf_{x \to x_0}$$? Infimum of x at the limit point?
No, the "infimum of x at x0" is x0!

Strictly speaking "lim inf" applies to sequences. Normally "lim inf xn" means the infimum of all subsequential limits. "lim inf f(x)", as x goes to x0 is the infinimum of all possible subsequential limits of {f(xn)} over all possible sequences {xn} converging to x0.

#### benorin

Homework Helper
An alternate definition of lower semicontinuity (from Real and Complex Analysis, by Walter Rudin) is $f:X\rightarrow \mathbb{R}$, where X is a topological space is lower semicontinuous if

$$\left\{ x:f(x)>\alpha\right\}\mbox{ is an open set in X, } \forall \alpha\in\mathbb{R}$$.

It's not a friendly definition, but it is equivalent. Upper semicontinuity is defined the same with "<" in place of ">".

Infimum of semicontinuous function

hello again,

Let $$f$$ be lower semicontinous function. Say the infimum of $$f$$ exists and that $$f(x^*) = \inf_{x \in \textup{dom}(f)} f(x)$$. Let $$\{x_k\}$$ be a sequence converging to $$x^*$$. Since $$f$$ is lower semicontinuous, so

$$\liminf_{k \to \infty} f(x_k) \geq f(x^*)$$.

I am having problem imagining how the sequence would be like. The only one I can think of is $$\{x^*, x^*, x^*, \ldots \}$$. Is this valid?

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#### mathwonk

Homework Helper
my favorite definition of upper semi continuous is that the value jumps up at individual points. e.g. the dimension of the kernel of a matrix of functions is upper semicontinuous, because the kernel can be bigger at points where the determinants of more submatrices vanish.

lower semi continuous is just the opposite: the value jumps down at points. so the dimension of the cokernel of a family of maps should do that i guess.

Sorry I am not able to comprehend your reply -- mainly due to my lack of understanding.

Anyway, I manage to clear my doubt now. Please ignore my silly 'sequence' in my last post.

However I've a new question. Supposing $$f, \{x_k\}$$ and $$x^*$$ are as defined in my last post. Since f is lower semicontinuous at $$x^*$$, hence

$$\liminf_{k \to \infty} f(x_k) \geq f(x^*) = \inf_{x \in \textup{ dom}(f)} f(x)$$.

This can be equivalently written as $$\lim_{k \to \infty} f(x_k) \geq f(x^*)$$. Is this true?

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#### mathwonk

Homework Helper
epsilon, schmepsilon, if f(x) = 1 for all x except x=0, and f(0) = 0, is f lower semicontinuous?

#### benorin

Homework Helper
Sure, characteristic functions of open sets are always LSC (Lower SemiContinuous).

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