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Chapter 13 Question 25g of Spivak's Calculus 4th Ed reads as follows:

Let $\mathcal L(x)$ be the length of the graph of $f$ on $[a,x]$, and let $d(x)$ be the length of the straight line segment from $(a,f(a))$ to $(x,f(x))$. Show that if $\sqrt{1+(f')^2}$ is integrable on $[a,b]$ and $f'$ is right continuous at $a$, then $\displaystyle \lim_{x\to a^+}\frac{\mathcal L(x)}{d(x)}=1$

Here is Spivak's proof, provided by the solution manual:

We are considering $$\lim_{x \to a^+}\frac{\displaystyle\int_a^x \sqrt{1+(f')^2}}{\sqrt{(x-a)^2+[f(x)-f(a)]^2}}.$$ By the Mean Value Theorem, $f(x)-f(a)=(x-a)f'(\xi)$ for some $\xi \in (a,x)$, and by the Mean Value Theorem for Integrals, the numerator is $(x-a)\sqrt{1+f'(\eta)^2}$ for some $\eta \in (a,x)$. So we are considering: $$\frac{(x-a)\sqrt{1+f'(\eta)^2}}{\sqrt{(x-a)^2+f'(\xi)^2(x-a)^2}}=\frac{\sqrt{1+f'(\eta)^2}}{\sqrt{1+f'(\xi)^2}},$$ which approaches $1$ as $x \to a^+$ since $\displaystyle \lim_{x \to a^+}f'(x)=f'(a)$.

I take issue with Spivak's proof because I believe we need more assumptions than Spivak is using in order for his argument to be valid. In particular, Spivak only requires that $f'$ is right continuous at $a$. However, I believe that in order to invoke the Mean Value Theorem for Integrals, we need to know that $f'$ is continuous in some neighborhood (not just a point) to the right of $a$. i.e. we need to assume that $f'$ is continuous on $[a,z]$ for some $z \in (a,b)$. Is this correct?


EDIT

Or is the idea that $f'$ must be implicitly defined for some interval $[a,z]$, in which case, the intermediate value property of the derivative (i.e. Darboux's Theorem) kicks in...and under such circumstances, we can 'get around' the fact that $f'$ is not assumed to be continuous on $[a,z]$?

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  • $\begingroup$ I don't understand what you mean by "in some neighbourhood of a", if it is continuous at $a$ then it is continuous for some sort of finite region around $a$, that is what that means. And we are taking the limit as the region shrinks to zero so we don't need to worry how big the actual region is. $\endgroup$ Commented Aug 23, 2022 at 6:12
  • $\begingroup$ @SuzuHirose are you suggesting that if a function is continuous at a point, then it is continuous in some neighborhood around a point? I think this is a false statement, classically demonstrated by Thomae's Function (en.wikipedia.org/wiki/Thomae%27s_function). Also, the issue is not with the limit - it is with the invocation of the Mean Value Theorem of Integrals. $\endgroup$
    – S.C.
    Commented Aug 23, 2022 at 6:18
  • $\begingroup$ Given any $\epsilon>0$ we can find $\delta>0$ such that if $0<x-a<\delta$ then $|f(x)-f(a)|<\epsilon$, so for any $\epsilon$ you like there is a non-zero $\delta$ within which $|f(x)-f(a)|<\epsilon$. $\endgroup$ Commented Aug 23, 2022 at 6:29
  • $\begingroup$ @SuzuHirose I am not sure I understand the relevance of your comment. You appear to have just restated the definition of continuity of $f$ at $a$. $\endgroup$
    – S.C.
    Commented Aug 23, 2022 at 6:31
  • $\begingroup$ I've posted an answer, let me know what you think. $\endgroup$ Commented Aug 23, 2022 at 6:57

2 Answers 2

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I see that you have a problem with applying intermediate value theorem for the integral in the numerator, basing only on the right-continuity at $a.$

In order to avoid that problem it suffices to show that $$\lim_{x\to a^+}{1\over x-a}\int\limits_a^x h(x)\,dx=h(a)$$ where $h$ ($h=\sqrt{1+f'^2}$) is Riemann integrable on some interval $[a,b]$ and right-continuous at $a.$ It can be done as follows. $$\left |{1\over x-a}\int\limits_a^x h(t)\,dt-h(a)\right |={1\over x-a}\left |\int\limits_a^x [h(t)-h(a)]\,dt\right |$$ By assumptions for any $\varepsilon>0$ there exists $\delta>0$ such that $$|h(t)-h(a)|<\varepsilon, \quad a\leq t<a+\delta$$ Let $a< x<a+\delta.$ Then $|h(t)-h(a)|<\varepsilon $ for $ a\leq t\leq x.$ Hence $$-\varepsilon<h(t)-h(a)<\varepsilon,\quad a\leq t\leq x$$ Integrating over $[a,x]$ and dividing by $x-a$ gives $$-\varepsilon < {1\over x-a} \int\limits_a^x [h(t)-h(a)]\,dt<\varepsilon,\quad a<x< a+\delta$$ Finally $${1\over x-a}\left |\int\limits_a^x [h(t)-h(a)]\,dt\right |<\varepsilon, \quad a<x< a+\delta$$

Remarks The proof can be made simpler if we assume continuity of $h.$ The right continuity of $\sqrt{1+(f')^2}$ can be replaced by the existence of the limit $\lim_{x\to a^+}\sqrt{1+(f')^2(x)}.$

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  • $\begingroup$ The theorem you prove is what one calls Fundamental theorem of calculus (for functions which are not necessarily continuous). Often the theorem is not stated this way but rather in terms of derivatives and integrals. +1 $\endgroup$
    – Paramanand Singh
    Commented Aug 23, 2022 at 9:12
  • $\begingroup$ @ParamanandSingh Thanks for the upvoting and the comment. I do not think it is FTC. The latter states that if $F$ is an antiderivative of $f$ and $f$ is Riemann integrable, then $$\int\limits_a^bf(x)\,dx=F(b)-F(a)$$ So the equality of $F'_+(a)=f(a)$ is assumed. In my answer it is not assumed, but proved. The proof of FTC is usually done by $F(b)-F(a)=\sum F(x_i)-F(x_{i-1})$ and applying the Lagrange theorem. There are no estimates. $\endgroup$ Commented Aug 23, 2022 at 9:39
  • $\begingroup$ The theorem in your last comment is second FTC. First FTC says that if $f$ is Riemann Integrable on $[a, b] $ and $f$ is continuous at $c\in[a, b] $ then $F'(c) =f(c) $ where $F(x) =\int_a^x f(t) \, dt$. $\endgroup$
    – Paramanand Singh
    Commented Aug 23, 2022 at 12:30
  • $\begingroup$ @ParamanandSingh Usually continuity in the first theorem is assumed and the mean value theorem is applied. See en.m.wikipedia.org/wiki/Fundamental_theorem_of_calculus $\endgroup$ Commented Aug 23, 2022 at 13:44
  • $\begingroup$ @RyszardSzwarc thank you for the answer. Do you think this is what Spivak intended on the reader inferring? I cannot make sense of the other post (Suzu Hirose's response)...so your interpretation is the only thing I can go off. Any idea what Suzu was trying to suggest? $\endgroup$
    – S.C.
    Commented Aug 24, 2022 at 7:46
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Say that we are considering the mean value between points $a$ and $b$ to be at the point $x$. Since $f'$ is right-continuous at $a$, for any $\epsilon>0$, choose $\delta$ so that $|f'(a)-f'(x)|<\epsilon/2$ and $|f'(a)-f'(b)|<\epsilon/2$. Then $$ |f'(x)-f'(b)|\leq|f'(a)-f'(x)|+|f'(b)-f'(a)|\\=\epsilon $$ so we have "enough" continuity in this region to guarantee that the mean value theorem is valid.

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