Let $p$ and $q$ be dual exponents and let $f \in L^p(0,1)$ and $g \in L^q(0,1)$ be non-zero functions. Further assume that $||fg||_1=||f||_p||g||_q$. Show that $f=cg^{q-1}$ where $c$ is constant.
By Hölder we know that $||fg||_1 \leq||f||_p||g||_q$, but I'm not sure how this helps us. By setting $f=cg^{q-1}$ I was able to verify the equality, but I don't know how to prove this is the only solution.
I will assume that the function are non-negative. Otherwise, the result may not hold, as we can have when $q$ is an even integer $f=1$ and $g=1$ on $(0,1/2)$ and $-1$ on $(1/2,1)$ or even worse $g^{q-1}$ may not be well-defined if $q$ is not rational and $g$ take negative values.
Since $f$ and $g$ are not zero, by scaling, it suffices to consider the case where $\left\lVert f\right\rVert_p=\left\lVert q\right\rVert_q=1$.
Observe that for all positive numbers $a$ and $b$, by convexity of the exponential function, $$ ab=\exp\left(\frac 1p\ln \left(a^p\right)+\frac 1q\ln \left(b^q\right)\right) \leqslant \frac 1p\exp\left(\ln \left(a^p\right) \right)+ \frac 1q\exp\left(\ln \left(b^q\right) \right) $$ hence $$ \tag{*} ab\leqslant \frac{a^p}p+\frac{b^q}q, \quad a,b\geqslant 0 $$ and by strict convexity, equality holds if and only if $a^p=b^q$. Let $$ h:=\frac{f^p}p+\frac{g^q}q-fg. $$ By $(*)$, $h$ is non-negative and by assumption, $\int h=0$ hence $h=0$ almost everywhere. By the equality case in $(*)$, we derive that $f^p=g^q$ almost everywhere.