let $(X,A,\mu)$ a measurable space.$1<p, q, s<\infty$
a.Prove that if $f\in L^p$ and $g\in L^q$ such that $1/p+1/q=1/s$ then $fg\in L^s$ and $\|fg\|_s\leq \|f\|_p\|g\|_q$.
b.Prove that if $f\in L^p$ , $g\in L^q$ and $h\in L^s$ such that $1/p+1/q+1/s=1$ then $fgh\in L^1(\mu)$ and $\|fgh\|_1\leq \|f\|_p\|g\|_q\|h\|_s$
My trial:
a. $\|fg\|_s= (\int_X|fg|^s)^{1/s}=(\int_X|fg|^s)^{1/p+1/q}= (\int_X |fg|^s)^{1/p} (\int_X |fg|^s)^{1/q}\leq$ (p,q>s) $(\int_X |fg|^p)^{1/p} (\int_X |fg|^q)^{1/q}= \|f\|_p\|g\|_q$ and since $\|f\|_p , \|g\|_q < \infty$ then $\|fg\|_s$ is finite too.so $fg \in L^s$
b. Denote $1/p+1/q=1/p'$ so by part a we get $fg\in L^{p'}$ and: $\|fg\|_{p'} \leq \|f\|_p \|g\|_q$, then we have $1/p'+1/s=1$ therefore by holder ineq1uality we get $\|fgh\|_1\leq \|fg\|_{p'} \|h\|_s \leq \|f\|_p\|g\|_q\|h\|_s$
Is what I solved right? Glad to her any notifications.
If I understand correctly, in part (a),
you have used that $p>s$ implies that $$ \int_X |F|^s \le \int_X |F|^p$$ (and similarly for $q>s$). This is not true. For example consider $X=[1,\infty)$ with the usual Lebesgue measure, and $f(x) = 1/x$. Then $f\in L^2(X)$ but not $L^1(X)$, so $$\infty=\int_1^\infty\frac1{x} dx \not\le \int_1^\infty \frac1{x^2}dx<\infty. $$
(a counterexample also in $L^1$ is $f(x)=\mathbf 1_{[1,N]}/x$, for some $N$ very large; $\mathbf 1_A$ is the indicator of $A$.)
You then further went on to write $\|fg\|_p \|fg\|_q = \|f\|_p \|g\|_q$. This is also not true, and can be seen by scaling considerations: if you multiply $f$ by $2$, the left hand side is multiplied by $4$, but the right hand side only by $2$. So the equality is false.
Instead, here's a big hint: try use the normal Holder's inequality with the exponents $1=\frac1{p/s}+\frac1{q/s}$.
Some other minor points: you should write $fg\in L^s$, not $\|fg\|_s\in L^s$; and $p'$ is normally reserved for the Hölder conjugate; so in part (b), perhaps use $s'$ instead. (The proof for (b) is correct.)