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Let $f,g\in C^1([a,b])$ with $a<b$ then prove that

$$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2\le \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$$

It smells like there is some mean value theorem going around. But I tried it as follows:

Indeed, it springs from mean value theorem that There exists $c_1,c_2\in (a,b)$ such that

$$\frac{f(b)-f(a)}{b-a} = f'(c_1)~~~ and ~~~~\frac{g(b)-g(a)}{b-a} = g'(c_2)$$

Then I have $$\frac{f(b)-f(a)}{b-a}+ \left(\frac{g(b)-g(a)}{b-a}\right)^2= f'(c_1)+(g'(c_2))^2\le \max_{t\in [a,b]}\{f'(t)\}+\max_{t\in [a,b]}\{(g'(t)^2)\}$$

Which is however not the required inequality.

Can anyone help? how can I improve this ?

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  • $\begingroup$ Either this is a very, very weird question, or else you're messing with symbols: the inequality is senseless as $\;g\;$ doesn't appear on the left hand...and twice you wrote the same! $\endgroup$
    – DonAntonio
    Apr 1, 2018 at 9:07

2 Answers 2

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\begin{align*} &\dfrac{f(b)-f(a)}{b-a}+\left(\dfrac{g(b)-g(a)}{b-a}\right)^{2}\\ &=\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\left(\int_{a}^{b}g'(t)\dfrac{dt}{b-a}\right)^{2}\\ &\leq\int_{a}^{b}f'(t)\dfrac{dt}{b-a}+\int_{a}^{b}(g'(t))^{2}\dfrac{dt}{b-a}\\ &\leq\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}\int_{a}^{b}\dfrac{dt}{b-a}\\ &=\max_{t\in[a,b]}\{f'(t)+(g'(t))^{2}\}. \end{align*}

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  • $\begingroup$ No comment Jessen inequality thanks $\endgroup$
    – Guy Fsone
    Apr 1, 2018 at 9:15
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Treat $\max$ as $\sup$, then apply then fact that $\sup(A+B)=\sup(A)+\sup(B)$, for $A,B$ be nonempty bounded subset of $\mathbb R$ and $A+B=\{a+b:a\in A,b\in B\}$

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    $\begingroup$ This is pure bluff, a false hint. -1 $\endgroup$ Apr 1, 2018 at 9:14
  • $\begingroup$ this answer is not appropriate to the question. can you use your statement to give an answer? $\endgroup$
    – Guy Fsone
    Apr 2, 2018 at 7:57
  • $\begingroup$ Moreover, your hint is pointless. note the following : There is a significant difference between $$ \max_{t\in [a,b]}\{f'(t)+(g'(t))^2\}$$ and $$ \max_{s,t\in [a,b]}\{f'(s)+(g'(t))^2\}$$ in which the latest incorporate your statement but does not fit the question itself. $\endgroup$
    – Guy Fsone
    Apr 2, 2018 at 8:29
  • $\begingroup$ @Medo read my last comment. you can see the difference. $\endgroup$
    – Guy Fsone
    Apr 2, 2018 at 8:31
  • $\begingroup$ @ Guy Fsone. Thank you for the remark. Indeed the question asks for the stronger (smaller) bound $\max_{t\in [a,b]}\{ f^{\prime}(t)+(g^{\prime}(t))^2\}$ not the weaker (larger) bound $$\max_{s,t\in [a,b]}\{ f^{\prime}(s)+(g^{\prime}(t))^2\}=\max_{s\in [a,b]}\{ f^{\prime}(s)\}+\max_{t\in [a,b]}\{(g^{\prime}(t))^2\}=$$ $\endgroup$
    – Medo
    Apr 4, 2018 at 9:31

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