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Suppose that $f$ is differentiable on $[0, 1]$, with $f(0) = 0$ and $f'(x) \geq m > 0$ for each $x \in [0, 1]$. Show that there is a subspace $J \subseteq [0, 1]$, with length greater than or equal to $\frac{1}{2}$, so that $f(x) \geq \frac{m}{2}$ for each $x \in [a, b]$.

Could you give me a hint how to solve the exercise?

Sammy Black
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evinda
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    What is $[a,b]$? How is it related to $J$? What do you mean by "subspace" here (as this is not standard usage)? Do you mean that $J$ is a subinterval of the form $[a,b]$ (with $0\le a<b\le 1$) perhaps? – Cameron Buie Dec 18 '13 at 22:31
  • I, too, am confused by subspace... how does a subspace have a length? – apnorton Dec 18 '13 at 22:55

6 Answers6

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Try applying the mean value theorem to the interval $[0, \frac{1}{2}]$.

John Hughes
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  • I tried it and got $f'(w)=2f(\frac{1}{2})\geq m$ .So,$f(\frac{1}{2})\geq\frac{m}{2}.$ But how can I find in that way the interval? – evinda Dec 19 '13 at 18:15
  • Well...$f$ is increasing, so from $1/2$ to $1$, f is greater than or equal to $m/2$, as other answers have explained as well. – John Hughes Dec 19 '13 at 19:52
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For every $x \in [0,1]$ we have $$ f(x)=\int_0^xf'(t)\,dt\ge \int_0^xm\,dt=mx. $$ Thus $$ f(x) \ge \frac12m \quad \forall x \in J:=[\frac12,1]. $$

HorizonsMaths
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Because $f'>m$ follows, that $f$ is monotonous increasing. So it is enough to show that $f(\frac12)>\frac m2 $. You can use the mean value theorem to prove this.

user116007
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What is the lowest possible value for $f(\frac 12)$?

Ragnar
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Define $g(x)=f(x)-mx$. Then $g'(x)=f'(x)-m>0$ so $g$ is strictly increasing. Therefore $g(x)\geq g(0)=0$ so $f(x) \geq mx$. From here you can guess the interval.

Beni Bogosel
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Is it not the case that

$f(x) = f(x) - f(0) = \int_0^x f'(t) dt \ge \int_0^x m dt = mx? \tag{1}$

Thus it is the case that

$f(x) \ge mx \tag{2}$

for $x \in [0, 1]$, whence for $ \in [1/2, 1]$ we have

$f(x) \ge (1 / 2)m; \tag{3}$

Now if we take the "subspace" $J$ to be the interval [1 / 2, 1]$, all the requisite conditions are satisfied. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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  • Yes,it was helpful...thank you!!! – evinda Dec 19 '13 at 18:10
  • @ evinda: you're welcome! Glad to be of service! You might consider upvoting/accepting answers which work for you; it help the answerer, obviously; and it helps you in that people will be more eager to provide you answers if that is the case. Happy Holidays! – Robert Lewis Dec 19 '13 at 18:17
  • Have also a nice holiday!!! – evinda Dec 19 '13 at 19:45
  • @ evinda: thanks for both the vote and the holiday blessings; and,once again, I wish you the same! – Robert Lewis Dec 19 '13 at 20:04