Continuity and differentiability for $g(x)$

Question

Below is my working:

CONTINUITY AT $x=a$

*$\lim_{x\to a^-}g(x)=\lim_{h\to0}g(a-h)=0$

*$\lim_{x\to a^+}g(x)=\lim_{h\to0}g(a+h)=\lim_{h\to0}\int_{a}^{a+h}f(t)dt=\int_{a}^{a}f(t)dt=0$

*$g(a)=\int_{a}^{a}f(t)dt=0$

Therefore $g(x)$ is continuous at $x=a$

CONTINUITY AT $x=b$

*$\lim_{x\to b^-}g(x)=\lim_{h\to0}g(b-h)=\lim_{h\to0}\int_{a}^{b-h}f(t)dt=\int_{a}^{b}f(t)dt$

*$\lim_{x\to b^+}g(x)=\lim_{h\to0}g(b+h)=\lim_{h\to0}\int_{a}^{b}f(t)dt=\int_{a}^{b}f(t)dt$

*$g(b)=\int_{a}^{b}f(t)dt$

Therefore $g(x)$ is continuous at $x=b$

CONTINUITY AT $x=k\in (a,b)$

*$\lim_{x\to k^-}g(x)=\lim_{h\to0}g(k-h)=\lim_{h\to0}\int_{a}^{k-h}f(t)dt=\int_{a}^{k}f(t)dt$

*$\lim_{x\to k^+}g(x)=\lim_{h\to0}g(k+h)=\lim_{h\to0}\int_{a}^{k+h}f(t)dt=\int_{a}^{k}f(t)dt$

*$g(k)=\int_{a}^{k}f(t)dt$

CONTINUITY AT $x=k<a$ and $x=k>b$

It can be shown in a similar way that $g(x)$ is continuous at $x=k<a$ and $x=k>b$

DIFFERENTIABILITY AT $x=a$

*$\lim_{x\to a^-}\frac{g(x)-g(a)}{x-a}=\lim_{h\to 0}\frac{g(a-h)-g(a)}{-h}=\lim_{h\to 0}\frac{0}{h}=0$

*$\lim_{x\to a^+}\frac{g(x)-g(a)}{x-a}=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=\lim_{h\to 0}\frac{\int_{a}^{a+h}f(t)dt-0}{h}=\lim_{h\to 0}\frac{\frac{d}{dh}\int_{a}^{a+h}f(t)dt}{1}=\lim_{h\to 0}f(a+h)=f(a)=0$

Therefore $g(x)$ is differentiable at $x=a$

DIFFERENTIABILITY AT $x=b$

*$\lim_{x\to b^-}\frac{g(x)-g(b)}{x-b}=\lim_{h\to 0}\frac{g(b-h)-g(b)}{-h}=\lim_{h\to 0}\frac{\int_{a}^{b-h}f(t)dt-\int_{a}^{b}f(t)dt}{-h}=\lim_{h\to 0}\frac{\frac{d}{dh}\int_{a}^{b-h}f(t)dt-0}{-1}=\lim_{h\to 0}f(b-h)=f(b)=\int_{a}^{b}f(t)dt$

*$\lim_{x\to b^+}\frac{g(x)-g(b)}{x-b}=\lim_{h\to 0}\frac{g(b+h)-g(b)}{h}=\lim_{h\to 0}\frac{\int_{a}^{b}f(t)dt-\int_{a}^{b}f(t)dt}{h}=\lim_{h\to 0}\frac{0}{h}=\lim_{h\to 0}0=0$

Therefore $g(x)$ is not differentiable at $x=b$

DIFFERENTIABILITY AT $x=k\in(a,b)$

*$\lim_{x\to k^-}\frac{g(x)-g(k)}{x-k}=\lim_{h\to 0}\frac{g(k-h)-g(k)}{-h}=\lim_{h\to 0}\frac{\int_{a}^{k-h}f(t)dt-\int_{a}^{k}f(t)dt}{-h}=\lim_{h\to 0}\frac{\frac{d}{dh}\int_{a}^{k-h}f(t)dt-0}{-1}=\lim_{h\to 0}f(k-h)=f(k)=\int_{a}^{k}f(t)dt$

*$\lim_{x\to k^+}\frac{g(x)-g(k)}{x-k}=\lim_{h\to 0}\frac{g(k+h)-g(k)}{h}=\lim_{h\to 0}\frac{\int_{a}^{k+h}f(t)dt-\int_{a}^{k}f(t)dt}{h}=\lim_{h\to 0}\frac{\frac{d}{dh}\int_{a}^{k+h}f(t)dt-0}{1}=\lim_{h\to 0}f(k+h)=f(k)=\int_{a}^{k}f(t)dt$

Therefore $g(x)$ is differentiable in $(a,b)$

DIFFERENTIABILITY AT $x=k<a$ and $x=k>b$

Similarly it can be shown that $g(x)$ is differentiable at $x=k<a$ and $x=k>b$

Am I correct? Is there any short way to solve this?

Answer 1

Yes, there is a much shorter way to do this; make use of the fundamental theorem of calculus! I didn't fully read through the details of your work, but it seems like you were almost rederiving the FTC (but you must have made a mistake somewhere). I'll state it here again for the sake of completeness:

FTC: Let $f:[a,b] \to \mathbb{R}$ be continuous, and define the function $g:[a,b] \to \mathbb{R}$ by \begin{equation} g(x) = \int_a^x f(t) \, dt \end{equation} Then, $g$ is differentiable on $[a,b]$ and $g'(x) = f(x)$.

Warning: at the endpoints $a$ and $b$, what we really mean is the right and left continuity of $f$ and differentiability of $g$. i.e \begin{equation} \lim_{h \to 0^+} f(a+h) = f(a) \end{equation} and \begin{equation} \lim_{h \to 0^+} \dfrac{g(a + h) - g(a)}{h} = f(a), \end{equation} and similarly at $b$, with $\lim_{h \to 0^-}$.

The correct answer is (A). First, lets see why $g$ is not differentiable at $a$. The left-derivative of $g$ at $a$ is $g_l'(a) = 0$ (the subscript $l$ means I'm taking the limit from the left of $a$), because to the left of $a$, $g$ is constant. Next, by the Fundamental Theorem of Calculus stated above, the right derivative is $g_r'(a) = f(a) \neq 0$. I said $\neq 0$, because the question states that the target space for $f$ is $[1, \infty)$. So, we showed that \begin{align} g_l'(a) = 0 \neq g_r'(a). \end{align} This proves $g$ is not differentiable at $a$. By similar reasoning, $g$ is not differentiable at $b$.

Edit:

I just read your proof for the continuity of $g$ at $a$. It has some mistakes. First, \begin{equation} \lim \limits_{x \to a^-}g(x) = \lim_{h \to 0} g(a-h) \end{equation} is an incorrect statement. It should be $\lim \limits_{h \to 0^+}$. The proper way to present it is \begin{equation} \lim \limits_{x \to a^-}g(x) = \lim_{x \to a^-} 0 = 0. \end{equation} Next, you wrote \begin{equation} \lim_{h \to 0} \int_a^{a+h} f(t) \, dt = \int_a^a f(t) \, dt = 0. \end{equation} There are a couple of mistakes here. Note that $f$ is only defined on $[a,b]$, so doing $\lim \limits_{h \to 0}$ does not make sense. You need to write $\lim \limits_{h \to 0^+}$. Next, why is \begin{equation} \lim_{h \to 0^+} \int_a^{a+h} f(t) \, dt = \int_a^a f(t) \, dt \end{equation} true? This is pretty much what you need to prove to say that "$g$ is continuous from the right at $a$"! So how would you go about fixing this?

Answer 2

Your solution uses Fundamental Theorem of Calculus implicitly (perhaps you are not aware where it is being used). Here are a few mistakes which you should know and then get rid of them:

1. The notation $\lim_{h\to 0}$ does not ensure that $h$ is positive. Use the notation $\lim_{h\to 0^+}$ instead.
2. The fact that $\lim_{h\to 0}\int_{a}^{k+h}f(t)\,dt=\int_{a}^{k}f(t)\,dt$ is just symbolic way of saying that $\int_{a} ^{x} f(t) \, dt$ is continuous at $k$. So your approach here does not prove the continuity of $g$ but rather assumes it. The result is true via Fundamental Theorem of Calculus for any Riemann integrable function $f$. Same remark applies to your handling of continuity at $a$ and $b$.
3. You have used L'Hospital's Rule for handling differentiability of $g$. Note that a derivative can't be calculated via L'Hospital's Rule. Rather derivatives are known by some other means and then used while applying L'Hospital's Rule. The key here is to use Fundamental Theorem of Calculus.

For more details on Fundamental Theorem of Calculus see this answer.