Identity between functions

Question

Let $f$ and $g$ be continuous functions of one real variable. We want to show that $\frac{\mathrm{d}}{\mathrm{d}t}f = g$ on the interval $[a,b]$. I have shown that for any subinterval $[t_a,t_b] \subseteq [a,b]$ we have $\int_{t_a}^{t_b} \frac{\mathrm{d}}{\mathrm{dt}} f(t) \, \mathrm{d}t = \int_{t_a}^{t_b} g(t) \, \mathrm{d}t$. Is it sufficient? Is there any variational way to justify this result?

Answer 1

Suppose $f^\prime(t)\neq g(t)$ for some $t\in(a,b)$ and assume without loss of generality that $f^\prime(t)-g(t) = D > 0$.

By continuity there exists an interval $(c,d)\ni t$ such that $f^\prime(s)-g(s)>\frac{D}{2}$ for all $s\in(c,d)$. Thus $\displaystyle \int_c^df^\prime(s)-g(s)\,\mathrm{d}s > \frac{D}{2}(d-c) >0$.

Hence, by contraposition, if $\displaystyle \int_{t_a}^{t_b} f^\prime(s)-g(s)\,\mathrm{d}s=0$ for any $[t_a,t_b]\subseteq[a,b]$, then $f^\prime(t)=g(t)$ for all $t\in(a,b)$.