# If $f(0)=0$ and $f''\ge 0$, then $f(a+b)\ge f(a)+f(b)$

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?

• Give him one more suit to alter!!. – Satish Ramanathan Jun 26 '15 at 18:14
• You cannot use Taylor series because you are not given that $f$ is analytic, among other issues – GFauxPas Jun 26 '15 at 18:15
• It is "Taylor" instead of "Tailor". In addition, the theorem has a remainder which cannot be thrown away at will. – Zhanxiong Jun 26 '15 at 18:16
• thanks, the remainders are still there, thats the expressions with c1 c2 c3 – AlphaBeta Jun 26 '15 at 18:19
• why would that be true? Think of a shifted parabola – 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.
• thanks for the reply. To clarify: doesn't $\ f$ need to be continuous in order to use the Mean Value Theorem? – AlphaBeta Jun 26 '15 at 18:33
• @Benwell Yes: $f$ needs to be both continuous and differentiable over the interval in question. – Jam Jun 26 '15 at 18:53
• @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. – Jam Jun 26 '15 at 18:57