Low value attained by real function with positive second derivative. Let $ f:[0,1]\to R$ be continuous with $f(0)=f(1)=0$. Let $f''(x)$ exist for all $x \in (0,1)$ with $\inf \{ f''(x) : x \in (0,1) \} =k>0 $. Prove that there exists $x \in (0,1)$ with $f(x) \le -k/8$. (I needed this as a tool to show by analytic means that the norm in an $l_p$ sequence space is strict. for $1<p< \infty$, that is, if $u,v$ are linearly independent then $||u||+||v|| >||u+v||$.)
 A: By the mean value theorem, we know that $f'(c)=0$ for some $c$. For $x>c$, another application of the mean value theorem yields $\frac{f'(x)-f'(c)}{x-c}=f''(p)$ for some $p$, and since $f''(p)\ge k$   we get $f'(x)\ge k(x-c)$. Similarly, for $x<c$ we obtain $f'(x)\le -k(c-x)$. Thus we have:
$$f'(x)\ge k(x-c)~~\text{if}~x>c\\
f'(x)=0~ ~\text{if}~~ x=c\\
f'(x)\le -k(c-x)~~\text{if}~x<c\\$$
Now divide the problem into two cases:
If $c>1/\sqrt 2$, note that for $x<c$ we have $f(x)=f(0)+\int_0^xf'(x)dx \le \int_0^x-k(c-x)dx=kx^2-kcx$. Now define $g(x)=kx^2-kcx+k/8$. Then $g(x)$ has a root $r$ in $(0,c)$ (here we use $c>1/\sqrt2$). Thus $f(r)\le kx^2-kcx=-\frac k8$, and the result is satisfied.
$$$$
If $c\le 1/\sqrt 2$, then we have $0=f(1)=f(c)+\int_c^1f'(x)dx\ge f(c)+\int_c^1k(x-c)dx$, so that $f(c)\le -\int_c^1f'(x)dx$. Now
$$-\int_c^1k(x-c)dx\le -\int_{\frac {1}{\sqrt 2}}^1k(x-c)dx=-k(\frac 12 -c +\frac {c}{\sqrt 2})$$, and since $c\le 1/\sqrt 2$ the last expression is less than $-\frac k8$, so the theorem is proved.
A: The derivative $f^{'}(x)$ may not exist at $x=0$ or $x=1$ but $M=\min \{f(x) : x \in [0,1] \}$ exists because  $f$ is continuous. Let $f(c)=M$ for some  $c \in [0,1]$ .We have  $0<c<1$, else $f$ is constantly $ 0$, contradicting  $f^{''} >0.$  And $f^{'}(c)=0$. Now for $x \in (0,1)$ and $x \ne c$, we have, for some $d$ between $c$ and $x$, $$f(x)=f(c)+(x-c)f^{'}(c)+(x-c)^2f^{''}(d)/2 $$$$= f(c) +(x-c)^2f^{''}(d)/2$$$$ \geq f(c)+(x-c)^2k/2.$$ Letting $x \to 0$ or letting $x \to 1$, we obtain, by continuity of $f$ at $0$ and $1$, $$0=f(0) \geq f(c)+c^2k/2$$ and $$0=f(1) \geq f(c)+(1-c)^2k/2$$. Therefore $$f(c) \leq   \min (-c^2k/2, -(1-c)^2/2 \le -k/8.$$
