Consider the function $f^2$, which is convex by given constraint. If there are two points $a$ and $b$ for which $f$ is zero, then it must be zero in $[a,b]$ by Jensen's inequality. Thus, $f=f'=f''=0$ in $(a,b)$. For proving the second derivative at the corners zero, use definition of derivative for $f'$ and evaluate the limit for $h\to 0^+$ for $a$ and $h\to 0^-$ for $b$ . The same approach works if $f$ is zero at exactly one point.

Let $A:=\{x:f(x)=0\}$ be the zero set of $f$ and suppose $a$ is a limit point of $A$. Hence $f(a)=0$ and we have some sequence $x_n\in A\setminus \{a\}$ with $x_n\rightarrow a$. By Rolle we have a sequence $y_n\in A\setminus \{a\}$ with $y_n\rightarrow a$ and $f'(y_n)=0$ for all $n$. Then $f'(a)=0$ and $f''(a)=\lim_n {{f'(y_n)-f'(a)}\over {y_n-a}}=0$. The rest is now standard: suppose $f''(a)\not=0$ for some $a\in I$. Then $f(a)=0$ and $a$ is an isolated root of $f$. Hence $f\not=0$ on a punctured nbhd $U:=\{0<|x-a|<\delta\}\subset I$. Then $f''=0$ on $U$ hence $f'$ is constant on $(a-\delta,a)$ and $(a,a+\delta)$. Since $f'$ is continuous on $(a-\delta,a+\delta)$, we find that $f'$ is constant on $(a-\delta,a+\delta)$. Hence $f''(a)=0$, contradiction.

Case 1: $f>0$ or $f<0$ for all $x$. In this case, the result is obvious. Case 2: There exist at least one point such that $f=0$. Sub-case 1: There exist $2$ or more such points: In this case let, two consecutive zeros be $a,b$, then $f(a)=f(b)=0$, WLOG let $f>0$ for all $x \in (a,b)$ then, by Rolle's, there exist $f'(c)=0$, but then for all $x \in (a,b)$ $f'',=0 \implies f'=k=0$. Thus $f$ is constant in $(a,b)$ and for all $x \in (a,b)$, $f(x)= \lim_{x \to a^+} f(x)=f(a)=0$ which is a contradiction. Sub-case 2: There exist only one point $d$ such that $f(d)=0$, notice, by similar argument, for all $x>d$ $f(x)=a_1 x+ b_1$ and for all $x<d$ $f(x)=a_2 x+ b_2$, and for the function to be continuous and differentiable at $d$, $a_1=a_2, b_1=b_2$, and so $f$ is a linear function and the result follows.