Let $a< f(x) < b $, $x \in \Omega $, $\mu(\Omega )=1 $, and set $t=\int f d \mu $.
Then $a < t < b $.
Suppose $\phi $ is a convex function on $(a,b) $
then by definition of convexity we have that for $a<s<t<u<b $, $\frac {\phi (t)- \phi(s)} {t-s } <\frac {\phi (u)- \phi(t)} {u-t } $
Define $\beta $ to be the supremum of the quotient on the left side.
From this it follows that $(u-t) \beta + \phi (t) \le \phi (u)$, $u \in (t,b) $.
Now Rudin seems to claim that the last inequality above is true more generally for $u \in (a,b) $
I could't verify this, so need help wiht the case $u \in (a,t) $.
Thanks in advance!
Since $$ \beta = \sup_{s\in(a,t)} \frac{\phi(t)-\phi(s)}{t-s} $$ we can take $s=u$ in the case $u\in(a,t)$ to get $$ \beta \ge \frac{\phi(t)-\phi(u)}{t-u} $$ Since $t-u>0$ in this case, this yields $$ (t-u)\beta \ge \phi(t)-\phi(u) $$ Rearranging yields the desired inequality.