Let $f: [0, \infty) \rightarrow \mathbb{R}$. Define the value of the left-hand limit of $f$ at $t>0$ to be $f(t^-) = \lim_{x \rightarrow t^-} f(x)$. Define the "left-hand limit function" of $f$ as $f^-: (0, \infty) \rightarrow \mathbb{R}, f^-(t) = f(t^-) = \lim_{x \rightarrow t^-} f(x)$. Prove that $f^-$ is left continuous at each $t>0$.

Let $t>0$. To prove $f^-$ is left continuous at $t$, I need to show that $\lim_{t \rightarrow t^-}f^-(t) = f^-(t) = f(t^-)$.

My question is, how can I use the sequential definition (not the $\epsilon-\delta$ definition) of left continuity to prove this question? In other words, let $(s_n)$ be a sequence contained in $(0, \infty)$ such that $s_n<t$ for all $n$ and $s_n \rightarrow t$. I need to show that $\lim_{n \rightarrow \infty} f^-(s_n) = f^-(t) = f(t^-)$. I have also been given a hint (but I have no idea where to incorporate it):

If $f(t^-)$ exists and is finite, then it is equivalent to: $$f(t^-) = \sup \inf_{s<t} \{f(v): s \le v < t\} = \inf \sup_{s <t} \{f(v) : s \le v < t\} $$

Any help would be greatly appreciated!

  • $\begingroup$ A working approach (which does not really use the hint): For every $s_n$, there is also a sequence $r_{n,k} \to s_n^-$ for $k\to\infty$. You can use this to pick a sequence $p_n$ such that each $p_n$ is equal to some $r_{n,k}$, we have $p_n \to t^-$ and each $f(p_n)$ is close to $f(s_n)$. $\endgroup$ – mlk Jul 27 '17 at 8:12

Using the definition of $f^-(t)$, given $\varepsilon>0$ there is $\delta>0$ such that $$|f(s)-f^-(t)|\le\varepsilon $$ for all $t-\delta\le s<t$. Since $s_n\to t^-$, there is $\bar n$ such that $t-\delta<s_n<t$ for all $n\ge \bar n$. Fix $n\ge \bar n$. Then for $t-\delta<s<s_n$ you have $$|f(s)-f^-(t)|\le\varepsilon $$ or equivalently, $$f^-(t)-\varepsilon\le f(s)\le f^-(t)+\varepsilon. $$ Letting $s\to s_n^-$ in the previous inequality, you get $$f^-(t)-\varepsilon\le f^-(s_n)\le f^-(t)+\varepsilon. $$ So you have $|f^-(s_n)- f^-(t)|\le \varepsilon$ for all $n\ge \bar n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.