Given $\ f$ so $\ f''(x) \ge 0$ for every $\ x \ge 0$, also $\ f(0)=0$.

Trying to show that if $\ a,b \ge 0 \Rightarrow f(a+b) \ge f(a) + f(b)$

Using Taylor I used $\ f(0)=0$ and got

$$f(x)=f'(0)x + \frac {f''(0)}2x^2$$

Plugging in$\ a, b$ and $\ a+b$ I got to this: $$f''(c3)(a+b)^2\ge f''(c1)a^2+f''(c2)b^2$$

What am I missing here? Is there a better way to solve this than Taylor?

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    $\begingroup$ Give him one more suit to alter!!. $\endgroup$ – Satish Ramanathan Jun 26 '15 at 18:14
  • $\begingroup$ You cannot use Taylor series because you are not given that $f$ is analytic, among other issues $\endgroup$ – GFauxPas Jun 26 '15 at 18:15
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    $\begingroup$ It is "Taylor" instead of "Tailor". In addition, the theorem has a remainder which cannot be thrown away at will. $\endgroup$ – Zhanxiong Jun 26 '15 at 18:16
  • $\begingroup$ thanks, the remainders are still there, thats the expressions with c1 c2 c3 $\endgroup$ – AlphaBeta Jun 26 '15 at 18:19
  • $\begingroup$ why would that be true? Think of a shifted parabola $\endgroup$ – Thomas Jun 26 '15 at 18:22

Wlog. $a\le b$. Then $f(a)-f(0)=af'(\xi)$ with $\xi\in(0,a)$ and $f(a+b)-f(b)=bf'(\eta)$ with $\eta\in(b,a+b)$ by the Mean Value Theorem. As $\xi<a\le b<\eta$ we have $f'(\eta)-f'(\xi)=(\eta-\xi)f''(\zeta)\ge0$ for some $\zeta\in (\xi,\eta)$ again by MVT. Therefore $$ f(a+b)-f(b)=bf'(\eta)\ge af'(\xi)=f(a)-f(0),$$ whence the claim.

  • $\begingroup$ thanks for the reply. To clarify: doesn't $\ f$ need to be continuous in order to use the Mean Value Theorem? $\endgroup$ – AlphaBeta Jun 26 '15 at 18:33
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    $\begingroup$ Differentiability implies continuity $\endgroup$ – user217285 Jun 26 '15 at 18:53
  • $\begingroup$ @Benwell Yes: $f$ needs to be both continuous and differentiable over the interval in question. $\endgroup$ – Jam Jun 26 '15 at 18:53
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    $\begingroup$ @Benwell If $f$ isn't continuous then you certainly can't use a Taylor Series for it. In the ranking of qualities of functions, so to speak, some functions are continuous; some of those are differentiable; some of those can be represented by Taylor Series. If Hagen's solution holds, yours does too. If $f$ isn't continuous, I don't think you can do much. $\endgroup$ – Jam Jun 26 '15 at 18:57

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