0

Which one of the following functions is left continuous and how to prove it? $$ f(x;\mathbf{d},\mathbf{m}) = x - \sum_{i=1}^{n}(x-d_{i})^{+}\mathbf{1}_{\{x<m_{i}\}}, $$ $$ g(x;\mathbf{d},\mathbf{m}) = x - \sum_{i=1}^{n}(x-d_{i})^{+}\mathbf{1}_{\{x\leq m_{i}\}}, $$ where, $\mathbf{b} = (b_1,\cdots,b_n) \in \mathbb{R}^{n}$, $\mathbf{m} = (m_1,\cdots,m_n) \in \mathbb{R}^{n}$ are given and $\mathbf{1}_{A} = 1$ if $A$ holds, otherwise $0$. The positive part function is $x^{+}= max(x,0)$.

Here, we only need to consider the left-continuity at $x = m_i, i = 1,\cdots,n.$

XWei
  • 39
  • What's the overscript plus sign? Also, which one do you think could probably be left continuous and why? – Niki Di Giano Aug 26 '18 at 07:39
  • Thank you for your interest. It is the positive part function sign, as $x^{+} = max(x,0)$. Maybe $f$ is l.c. But I am not sure and I hope there could be a proof. – XWei Aug 26 '18 at 08:16

1 Answers1

0

By the definition of left continuity: $$\lim_{x\to x_0^-} f(x) = f(x_0)$$

The function $f$ in your case has some value when it comes from the left, but suddenly jumps to another value as $x=m_i$.

Consider for example the biggest $m_i$, say $m_j$. The function $f$ jumps from some value (depending on $d_j$) to $m_j$, while $g$ does not.

So $g$ is left continuous, $f$ is not.

EDIT: even if the answer has been accepted, I'll give a hint for what the proof should look like.

For convenience, suppose the elements of $m$ are ordered like $m_1 < ... <m_n$. By linearity of the limit: $$\lim_{x\to m_1^-} f(x) = m_1 - (m_1 - d_1)^+ \\ f(m_1) = m_1$$ This fact alone proves that $f$ is not continuous from the left in general. $$\lim_{x\to m_1^-} g(x) = m_1 - (m_1 - d_1)^+ \\ g(m_1) = m_1- (m_1 - d_1)^+\\ \lim_{x\to m_2^-} g(x) = m_2 - \big( (m_2 - d_1)^+ + (m_2 - d_2)^+ \big) \\ g(m_2) = m_2 - \big( (m_2 - d_1)^+ + (m_2 - d_2)^+ \big)$$ And the same goes on for all other $m_i$. The key fact is that $1_A$ has a $\leq$ rather than a $<$ in its $A$ proposition.

Niki Di Giano
  • 1,708
  • How to prove it ? – XWei Aug 26 '18 at 10:25
  • Plugging in the definition values for $x_0 = m_i$ and simplifying will do. Topologically, left-continuous functions are continuous functions from $\Bbb{R}_u$ (the real line in the upper limit topology) to $\Bbb{R}$, you can see for yourself that preimages of open sets under $g^{-1}$ are unions of intervals of the form $(a, b)$ and $(a, b]$, which are open in $\Bbb{R}_u$. The same cannot be said of the function $f$. – Niki Di Giano Aug 26 '18 at 10:55
  • I think you are right. It seems that $g$ is l.c. But I still cannot prove that. – XWei Aug 26 '18 at 12:14
  • I have outlined a proof in my last edit. – Niki Di Giano Aug 26 '18 at 12:26
  • Thank you so much for your help. – XWei Aug 26 '18 at 13:12