Lowe Semicontinuity (1 Viewer)

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Lower Semicontinuity

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

[tex]\liminf_{x \to x_0} f(x) \geq f(x_0).[/tex]

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

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

[tex]\liminf_{x \to x_0} f(x) \geq f(x_0).[/tex]

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 [tex]f(x) > f(x_0) - \epsilon[/tex] 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 [tex]lim f(x_n)\geq f(x_0)[/tex] and so of course must lim inf.

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

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!
 
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HallsofIvy said:
Looks pretty straight forward to me. Suppose [tex]f(x) > f(x_0) - \epsilon[/tex] 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 [tex]lim f(x_n)\geq f(x_0)[/tex] and so of course must lim inf.!
This is the part I don't understand. Suppose [tex]f(x) > f(x_0) - \epsilon[/tex] for all x in some neighborhood U. Then for some x in the neighborhood of U, [tex]f(x) < f(x_0)[/tex] may hold true since [tex]\epsilon >0[/tex]. I am lost.

HallsofIvy said:
Conversely suppose [tex]\liminf_{x \to x_0} f(x) \geq f(x_0)[/tex] and suppose there were no neighborhood U as above. Let Un be (x0- 1/n, x0+ 1/n). Since none of these can satisfy [tex]f(x) > f(x_0) - \epsilon[/tex] for all x in Un, there must exist xn in Un such that [tex]f(x) \leq f(x_0) - \epsilon[/tex]. But then, for that sequence, [tex]lim f(x_n)\leq f(x_0)[/tex], contradicting [tex]\liminf_{x \to x_0} f(x) \geq f(x_0)[/tex].
I agree with this one.
 
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Oh I see my problem now. The keyword that I missed was "for every [tex]\epsilon > 0[/tex], ...." .
I have one last question. How does one read [tex]\liminf_{x \to x_0}[/tex]? Infimum of x at the limit point?
 

HallsofIvy

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kaosAD said:
Oh I see my problem now. The keyword that I missed was "for every [tex]\epsilon > 0[/tex], ...." .
I have one last question. How does one read [tex]\liminf_{x \to x_0}[/tex]? 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

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

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

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

hello again,

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

[tex]\liminf_{k \to \infty} f(x_k) \geq f(x^*)[/tex].

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

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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.
 
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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 [tex]f, \{x_k\}[/tex] and [tex]x^*[/tex] are as defined in my last post. Since f is lower semicontinuous at [tex]x^*[/tex], hence

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

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

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

benorin

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

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