- #1
kaosAD
- 33
- 0
I have questions regarding this subject.
By definition, [tex]\limsup_{k \to \infty} f(x_k) \equiv \lim_{k \to \infty} \sup_{n \geq k} f(x_n)[/tex] and [tex]\liminf_{k \to \infty} f(x_k) \equiv \lim_{k \to \infty} \inf_{n \geq k} f(x_n)[/tex]. Say a sequence [tex]\{x_k\}[/tex] converging to [tex]0[/tex] from the left in the following example.
[tex]f(y) = \left\{
\begin{array}{ll}
y + 1 & \quad ,y > 0 \\
y & \quad ,y \leq 0
\end{array}
\right.[/tex]
Then [tex]\limsup_{k \to \infty} f(x_k) = \liminf_{k \to \infty} f(x_k) = f(0)[/tex].
Suppose we have another sequence [tex]\{x_k\}[/tex] converging to [tex]0[/tex] from the right. Then [tex]\limsup_{k \to \infty} f(x_k) = \liminf_{k \to \infty} f(x_k) > f(0)[/tex].
What is the difference between [tex]\limsup_{k \to \infty} f(x_k)[/tex] and [tex]\liminf_{k \to \infty} f(x_k)[/tex]? I don't see any difference.
By definition, [tex]\limsup_{k \to \infty} f(x_k) \equiv \lim_{k \to \infty} \sup_{n \geq k} f(x_n)[/tex] and [tex]\liminf_{k \to \infty} f(x_k) \equiv \lim_{k \to \infty} \inf_{n \geq k} f(x_n)[/tex]. Say a sequence [tex]\{x_k\}[/tex] converging to [tex]0[/tex] from the left in the following example.
[tex]f(y) = \left\{
\begin{array}{ll}
y + 1 & \quad ,y > 0 \\
y & \quad ,y \leq 0
\end{array}
\right.[/tex]
Then [tex]\limsup_{k \to \infty} f(x_k) = \liminf_{k \to \infty} f(x_k) = f(0)[/tex].
Suppose we have another sequence [tex]\{x_k\}[/tex] converging to [tex]0[/tex] from the right. Then [tex]\limsup_{k \to \infty} f(x_k) = \liminf_{k \to \infty} f(x_k) > f(0)[/tex].
What is the difference between [tex]\limsup_{k \to \infty} f(x_k)[/tex] and [tex]\liminf_{k \to \infty} f(x_k)[/tex]? I don't see any difference.
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