$f : (a, b] \to \mathbb{R}$, where $f'(x)$ exists and bounded in $(a, b]$. Prove: $\lim_{x \to a^{+}}f(x)$ exists.

Question

Given that $f: (a, b] \to \mathbb{R}$, and the derivative $f'(x)$ exists and is bounded in $(a, b]$. Prove that $\lim_{x \to a^{+}}f(x)$ exists.

I attempt to use a contradiction by assuming that $\lim_{x \to a^{+}}f(x)$ does not exist. In this case there exists some $\epsilon > 0$ such that for all $\delta > 0$, there exists $x > a$ with $x - a < \delta$ and $|f(x) - f(a)| > \epsilon$. In particular we have some sequence $a_n \to a$ from the right-hand side and yet $f(a_{n})$ does not converge to anything. We also have that $f'$ exists and is bounded on $(a, b]$, and so for any $c$ in $(a, b]$, we have that $\lim_{x \to c}\frac{f(x) - f(c)}{x - c}$ exists. We also have that since $f'(x)$ exists, we have that for all $[c, b]$ with $c > a$, that $f(x)$ is uniformly continuous in $[c, b]$.

It is at this point that I am not sure what to contradict: either the existence of the derivative somewhere, or the fact that it is bounded. In one sub-case it seems intuitively clear that (e.g.) if the right-hand limit $\lim_{x \to a^{+}}f(x) = \infty$, then so should the derivative $f'(c)$ as $c \to a$, but this is just the sub-case where it goes to infinity. The main obstacle here is that $f(a)$ is not defined, and there is no obvious way to complete it using the right-handed limit. Any hints would be appreciated.

Edit: Currently I have this idea of a proof, using the idea of accumulation points (but doesn't need contradiction): We suppose above that $f'(x)$ exists in $(a, b]$. In that case, $f(x)$ is continuous in $(a, b]$, and uniformly continuous in $[c, b]$ for every $c$ in $(a, b]$. By the limit definition of continuity, we then have that the right-handed limit exists at every such point $c$. We should have that the set $\{f(x)\}$ for $x$ in $(a, b]$ is bounded from above. (This follows from MVT, since otherwise the average rate of change would get higher and higher, fixing one of the endpoints.) Then consider the point $a$, and any sequence $a_n \to a$ from the right-hand side, so each $a_n > a$. Then consider the sequence $f(a_n)$, and we have that $\{f(a_{n})\}$ is bounded from above since each $a_n$ is in $(a, b]$. Then by Bolzano Weierstrass we have an accumulation point, denoted "$f(a)$". Then if we take an arbitrary sequence $b_n \to a$, we should have $f(b_n) \to f(a)$, by using the triangle inequality. Is this idea sufficient?

Answer 1

You can proceed in your method, and discuss it in two cases. Since you assume $\lim_{x\to a^+} f(x)$ doesn't exsit.

Case.(1) $f(x)$ is unbounded, then you can find a sequence $a_n$, such that $a_n\to a^+$ and $f(a_n)\to \infty$. Then how to show this contradicts with the fact $f'(x)$ is bounded? (Hint below)

Choose a fixed point $x_0$, $|a_n-x_0|\le x_0-a$, but $|f(a_n)-f(x_0)| $ can be arbitrarily large.

Case.(2) $f(x)$ is bounded, then you can find two sequences, $x_n, y_n$, such that $x_n\to a^+, y_n\to a^+$, $\lim_{n\to \infty} f(x_n)=c_1, \lim_{n\to \infty} f(y_n)=c_2$ and $c_1\neq c_2$. Define the distance $d=c_1-c_2$. Can you find the contradiction with the fact $f'(x)$ is bounded? (Hint below)

Note that $d$ is some fixed number, but $|x_n-y_n|$ can be arbitrarily small.

Update:

Answer to your new idea:

Edit: Currently I have this idea of a proof, using the idea of accumulation points (but doesn't need contradiction): We suppose above that $f'(x)$ exists in $(a, b]$. In that case, $f(x)$ is continuous in $(a, b]$, and uniformly continuous in $[c, b]$ for every $c$ in $(a, b]$. By the limit definition of continuity, we then have that the right-handed limit exists at every such point $c$. We should have that the set $\{f(x)\}$ for $x$ in $(a, b]$ is bounded from above. (This follows from MVT, since otherwise the average rate of change would get higher and higher, fixing one of the endpoints.) Then consider the point $a$, and any sequence $a_n \to a$ from the right-hand side, so each $a_n > a$. Then consider the sequence $f(a_n)$, and we have that $\{f(a_{n})\}$ is bounded from above since each $a_n$ is in $(a, b]$. Then by Bolzano Weierstrass we have an accumulation point, denoted "$f(a)$". Then if we take an arbitrary sequence $b_n \to a$, we should have $f(b_n) \to f(a)$, by using the triangle inequality. Is this idea sufficient?

Your new idea only handles the unbounded case for $f(x)$. But for the bounded case, your current sketched idea still does not work, because by Bolzano-Weierstrass, there exsits an accumulation point, true. BUT, it may exist more than one accumulation point, how can you rule out other accumulation points?

If you assume there are more than one accumulation points, namely, there exists two sequences, $x_n, y_n$, such that $x_n\to a^+, y_n\to a^+$, $\lim_{n\to \infty} f(x_n)=c_1, \lim_{n\to \infty} f(y_n)=c_2$ and $c_1\neq c_2$. This will go back to my second hint.

Answer 2

If you want, you can obtain the result without using contradiction.

Let $M=\sup\limits_{x \in (a,b]} \lvert f^\prime(x) \rvert$. According to Mean Value Theorem, you have $\lvert f(x)-f(y) \rvert \le M \lvert x-y \rvert$ for any $x,y \in (a,b]$.

You can then leverage this inequality to prove that for any sequence $\{a_n\}$ converging to $a$, the sequence $\{f(a_n)\}$ is Cauchy and therefore converges, which leads to the desired result using sequential continuity.

Answer 3

Choose $M>0$ such that $|f'(x)|\leq M$ for all $x\in(a,b)$. Let $(x_{n})$ be an arbitrary sequence in $(a,b)$ such that $x_{n}\rightarrow a$. We go to prove that $\left(f(x_{n})\right)_{n}$ converges by showing that it is a Cauchy sequence. Let $\varepsilon>0$. Choose $N$ such that $|x_{n}-x_{m}|<\frac{\varepsilon}{M}$ whenever $m,n\geq N$. Let $m,n\geq N$, then there exists $\xi_{m,n}$ between $x_{n}$ and $x_{m}$ such that \begin{eqnarray*} |f(x_{n})-f(x_{m})| & = & \left|f'(\xi_{m,n})(x_{n}-x_{m})\right|\\ & \leq & M\cdot|x_{n}-x_{m}|\\ & < & \varepsilon. \end{eqnarray*}

Next, we go to show that $\lim_{n\rightarrow\infty}f(x_{n})$ is independent of the choice of $(x_{n})$. Let $(x_{n})$ and $(y_{n})$ be sequences in $(a,b)$ such that $x_{n}\rightarrow a$ and $y_{n}\rightarrow a$. For each $n,$ there exists $\eta_{n}$ between $x_{n}$ and $y_{n}$ such that $f(x_{n})-f(y_{n})=f'(\eta_{n})(x_{n}-y_{n})$. We have that \begin{eqnarray*} \left|f(x_{n})-f(y_{n})\right| & = & \left|f'(\eta_{n})(x_{n}-y_{n})\right|\\ & \leq & M|x_{n}-y_{n}|\\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty$. Therefore, $\lim_{n\rightarrow\infty}f(x_{n})=\lim_{n\rightarrow\infty}f(y_{n})$. Denote the common limit by $L$. By Heine Theorem, $\lim_{x\rightarrow a+}f(x)=L$.