I have a long and complicated function $f(T)$ whose derivative is not easy to derive. So, I want to prove that $f(T)$ is a convex function over the interval $(0,\infty)$ . I think it is easier by using contradiction. The convexity definition of a function states that $f(T)$ is convex over an interval $(a,b)$ if for every $T_1, T_2 \in (a,b)$, $0\leq \lambda \leq 1$ and the inequality below holds
$f\left( \lambda T_1 + (1-\lambda)T_2 \right) \leq \lambda f(T_1) + (1-\lambda)f(T_2)$.
Now, I assume that for every $T_1, T_2 \in (0,\infty)$ and $0\leq \lambda \leq 1$, the inequality below is true:
$f\left( \lambda T_1 + (1-\lambda)T_2 \right) > \lambda f(T_1) + (1-\lambda)f(T_2)$.
Then, let $T_1 = T_2 = 1$ and $\lambda = 1/3$. The inequality becomes $f(\frac{1}{3} + \frac{2}{3}) > \frac{1}{3}f(1) + \frac{2}{3}f(1)$ $f(1) > f(1)$ which is wrong. By contradiction, $f(T)$ is convex. But, then every function can be proven to be complex. How should I fix this proof?
If $f$ is not convex you can only say that $f(\lambda T_1+(1-\lambda )T_2)>\lambda f(T_1)+(1-\lambda )f(T_2))$ for some choices of $T_1,T_2,\lambda $, not for all choices.