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I found this question on the 1998 Roorkee Paper during my crazy hunt for mind-blowing calculus questions.

Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function, differentiable on $(0,1)$ such that $\int_0^{a} f(x) dx=0$ for some $a \in(0,1)$, then

(1) $\left|\int_0^1 f(x) d x\right| \leq \frac{1-a}{2} \max _{0<x<1}\left|f^{\prime}(x)\right|$

(2) $\left|\int_0^1 f(x)dx\right|>\frac{1-\mathrm{a}}{2} \max_{0<x<1}\left|f'(x)\right|$

(3) $\left|\int_0^1 f(x) d x\right|=\frac{1-a}{2} \max_{0<x<1}\left|f^{\prime}(x)\right|$ for some function $f(x)$

(4) none of the other options is correct

And ofcourse I have no idea on how to solve this. However on carefully reading the options and noticing $\frac{1-a}{2}$, I thought of making a connection with the concept of 'Definite Integral as Limit of Sum' where $\frac{b-a}{h}$ could be $\frac{1-a}{2}$ here.

Then I also believe there is an involvement of Lagranges Mean Value Theorem somewhere as the questions says continuous and differentiable.

I presented my observations to my teacher and he gave me a hint: "Put $x=at$ and change limits accordingly and try to apply your concepts."

Could someone kindly help me with this one.

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  • $\begingroup$ Can you find counter-examples to two of the possible answers? $\endgroup$
    – Henry
    Jun 1, 2023 at 13:16
  • $\begingroup$ @Henry Didnt get you $\endgroup$ Jun 1, 2023 at 13:42
  • $\begingroup$ I suggest you play around with simple functions $f$, such as $f(x) = x - c$ for various constants $c$. Can you choose $c$ so that (1) is false, and or so that (2) is false, or so that (3) is true? Note that the value $a$ in the problem statement will be different for each $c$. If this doesn't work, move on to somewhat more complicated functions. Even if it doesn't solve your problem by itself, it will give you some idea of what is going on here, which can help you move forward. $\endgroup$ Jun 3, 2023 at 1:08
  • $\begingroup$ I took that Roorkee entrance exam in 1998, but I am not able to recall this question at all. The integral $\int_0^1 f(x) \, dx$ can be simply replaced by $\int_a^1 f(x) \, dx$ which equals $(1-a)f(c)$ for some $c\in(a, 1)$ and then we need to compare values of $f(c) $ with that of $2f'(x)$. $\endgroup$
    – Paramanand Singh
    Jun 6, 2023 at 7:31
  • 2
    $\begingroup$ Side note: it's not like you're doing anything wrong, but I feel like the questions you've been asking on this site are too advanced for your current level. It's like if a student, who's learning what sums are and what $\sum$ means, asks questions that involve Taylor series. I've been noticing that whenever you get an answer, you still have questions about fundamental concepts important to your OPs. Of course, curiosity is good and I don't know where your experience lies in math, but I suggest reading a math textbook about the basics of proofs to gain some self-sufficiency. $\endgroup$ Jun 8, 2023 at 10:45

3 Answers 3

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Let $M=max_{0<x<1} |f’(x)|$.

This is tricky. Let me give some intuition first. It’s best to think of $M$ as constant and then trying to maximize the integral. We know that after $a$, $f$ is linearly bounded, so we need to try to get a good upper bound for $f(a)$. Since the derivative of $f$ is bounded, we need to try to show that $f$ itself can’t be too large. One way to relate an integral with derivatives is integration by parts which happens to work out well. On to the proof.

Applying integration by parts to $x f(x)$ gives:

$$af(a)=\int^a_0 f(x)dx + \int^a_0 xf’(x)dx$$ $$af(a)=\int^a_0 xf’(x)dx$$ $$|af(a)|\leq \int^a_0 x|f’(x)| dx \leq M \int^a_0 xdx=M a^2/2$$

Thus, $|f(a)|\leq M a/2$, so for $a<x$, $|f(x)|\leq |f(a)|+|f(x)-f(a)| \leq Ma/2+M(x-a)$.

$$|\int_0^1 f(x)dx|= |\int^1_a f(x)dx|$$ $$ \leq \int^1_a |f(x)|dx$$ $$ \leq \int^1_a (Ma/2+M(x-a))dx$$ $$ \leq M\int^1_a (x-a/2)dx$$ $$ \leq M(1^2/2-a^2/2-a/2(1-a))dx$$ $$ \leq M(1-a)/2$$

Thus, the answer is (1).

Note that if you haven’t gotten around to integration by parts yet, there’s another way to get a bound on $f(a)$. Consider $f(x)-M(x-a/2)$. It has nonpositive slope and integrates to $0$ from $0$ to $a$. By the mean value theorem, it must be 0 at some point and since it has nonpositive slope, we get that it’s final value is negative, so $f(a) - M(a-a/2) \leq 0$, so $f(a) \leq Ma/2$. The same argument on $-f(x)$ gives that $-f(a) \leq Ma/2$, so $|f(a)|\leq Ma/2$.

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$$\int_0^afdx=0=xf\bigg|_0^a-\int_0^a xf'dx=af(a)-\int_0^a xf'dx$$

Let $\displaystyle m=\max_{x\in(0,1)} |f'|$, we get

$$|af(a)|=\left|\int_0^a xf'dx\right|\le m \left|\int_0^a xdx\right|=m\frac12a^2\Longrightarrow |f(a)|\le \frac a2m$$

hence

$$\begin{align}\left|\int_0^1 fdx\right|&=\left|\int_a^1 fdx\right|=\left|\int_a^1 f(a)+f'(\xi )(x-a)dx\right|\\ \\ &\le\left|\int_a^1 f(a)dx\right|+\left|\int_a^1f'(\xi )(x-a)dx\right|\\ \\ &\le\left|\int_a^1 f(a)dx\right|+m\left|\int_a^1(x-a)dx\right|\\ \\ &= (1-a)|f(a)|+m\cdot\frac12(1-a)^2\\ \\ &\le (1-a)\frac a2m+m\cdot\frac12(1-a)^2\\ \\ &=\frac{1-a}2m\\ \\ &=\frac{1-a}2\max_{x\in(0,1)} |f'|\end{align}$$

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The conclusion of $(1)$ and $(3)$ follow straightforward from the Taylor formula applied to the antiderivative $F(x)=\int\limits_0^xf(t)\,dt.$ We have $F(1)=\int\limits_0^1f(t)\,dt$ and $$M:=\max_{0\le x\le 1} |f'(x)|=\max_{0\le x\le 1} |F''(x)|$$ By assumption $F(a)=0$ for some $0<a<1.$ The Taylor formulas at $0$ and $1,$ with respect to $a,$ yield $$ 0=F(0)=F'(a)(-a)+{F''(\eta_1)\over 2}a^2\\ F(1)=F'(a)(1-a)+{F''(\eta_2)\over 2}(1-a)^2$$ for some $0<\eta_1<a<\eta_2<1.$ Calculating $F'(a)$ from the first equality and substituting it into the second one gives $$F(1)={F''(\eta_1)\over 2}a(1-a)+{F''(\eta_2)\over 2}(1-a)^2\quad (*)$$ Therefore $$|F(1)|\le {M\over 2}a(1-a)+{M\over 2}{(1-a)^2} ={M(1-a)\over 2}\quad (**)$$ In view of $(*),$ the equality in $(**)$ occurs when $F''=f'$ is constant, i.e. $f$ is an affine function vanishing at $x_0,$ $0<x_0<{1\over 2},$ for example $f(x)=3x-1.$ Clearly it also holds for the trivial case $f\equiv 0.$

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