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$f : [0,1] \to \mathbb R$ is continuous, and $f(0) = f(1) = 0$. $f''$ exists and $f''(x) ≥ 0$ at all $x \in (0,1)$. Show that $f(x) \le 0$ for all $x \in (0,1)$.

My idea:

  1. Use Rolle's Theorem, there exists point $c$ such that $f'(c)=0$.
  2. Since $f''(x) \ge 0$, then $f'(x)$ strictly increasing.
  3. By 1 + 2, $f$ attains minimum at $c$. (How to show that ?)
  4. Stuck here, though I can visualize the graph in my mind. By contradiction?
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  • $\begingroup$ Rolle's theorem only implies there is at least one point $c$ where $f'(c)=0$. There could be far more than one. It can happen that only one of those is where $f$ attains its minimum. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 1 '14 at 5:11
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I tried this :(please check)

Suppose $f(x)>0$ for some $x\in (0,1)$.Then $\exists a\in (0,1) $ such that $f(a)>0$

Consider $[a,1]$ By MVT $\exists c \in (0,1)$ such that

$\frac{f(1)-f(a)}{1-a}=f^{'}(c)<0$

Consider $[0,a]$ By MVT $\exists d \in (0,a)$ such that

$\frac{f(a)-f(0)}{a-0}=f^{'}(d)>0$

Now $f^{''}(x)>0 \implies f^{'}(x)$ is increasing

But here $d<c \implies f^{'}(d)>f^{'}(c)$ contradiction to $f^{'}(x)$ is increasing

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  • $\begingroup$ very simple and very clear +1. $\endgroup$ – Paramanand Singh Dec 1 '14 at 8:30
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I think contradiction works better here.

Hint: suppose (for contradiction) that there is some $a \in (0,1)$ such that $f(a) > 0$. It follows by the mean value theorem that there exists a $c \in (0,a)$ such that $f'(c) = \frac{f(a) - f(0)}{a-0} > 0$.

Now, $f''\geq 0$ on $(c,1)$, which contains $(a,1)$. What do we conclude?

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  • $\begingroup$ What can we say about $f(1$? Try drawing a picture. $\endgroup$ – Omnomnomnom Dec 1 '14 at 5:40
  • $\begingroup$ I have added an answer below .Please check $\endgroup$ – Learnmore Dec 1 '14 at 7:29

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