Show if $x\in l^p$ and $y\in l^q$ with $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$ then $\|xy\|_r\leq\|x\|_p\|y\|_q$ Show if $x\in l^p$ and $y\in l^q$ with $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$ then $\|xy\|_r\leq\|x\|_p\|y\|_q$. My intuition is to use Young's Inequality and then apply it to $A_k=\frac{|x_k|}{\|x\|}$ and $B_k=\frac{|y_k|}{\|x\|}$. I can't seem to get the required result though. Any advice?
 A: I'll prove it directly for an arbitrary measure space $({\cal X}, A, \mu)$. It is not the most efficient way to do it - you can use Hölder's Inequality, but I think that this proof makes sure the idea hits home.
Let $a,b \in \Bbb R^+$, e $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$. So: $$(ab)^r \leq r \left(\frac{a^{p}}{p} + \frac{b^{q}}{q}.\right)$$
Note that $r = \frac{pq}{p + q}$. Consider the function $f: \mathbb{R} \to \Bbb R$ given by $f(x) = e^x$. We have that $f''(x)  = e^x> 0$, and hence $f$ is convex. So given $a,b \in \Bbb R$, and $t \in [0,1]$, we get: $$f(tx + (1-t)y) \leq tf(x) + (1-t)f(y),$$ that is: $$e^{tx}e^{(1-t)y} \leq te^x + (1-t)e^y.$$ Since $a,b > 0$, we can choose $x = \ln a^{p}$ e $y = \ln b^{q}$. More, we can choose $t = \frac{q}{p+q}$. Observe that: $$1-t = 1 - \frac{q}{p+q} = \frac{p}{p+q}.$$
Substituiting, we obtain:
$$a^rb^r \leq \frac{qa^{p} + p b^{q}}{p + q} = \frac{\frac{p}{p}qa^{p} + \frac{q}{q}p b^{q}}{p+q} = r\left(\frac{a^{p}}{p} + \frac{b^{q}}{q}\right).$$
Now we'll use this stronger Young's Inequality to go on.
Make $$a = \frac{|f(x)|}{\|f\|_{p}} \quad\mbox{and}\quad b = \frac{|g(x)|}{\|g\|_{q}}$$ above. So: $$\frac{|f(x)g(x)|^r}{\|f\|_{p}^r \|g\|_{q}^r} \leq \frac{r}{p} \frac{1}{\|f\|_{p}^{p}} |f(x)|^{p} + \frac{r}{q} \frac{1}{\|g\|_{q}^{q}} |g(x)|^{q}$$
Integrate: $$\frac{1}{\|f\|_{p}^r\|g\|_{q}^r}\int_{\mathcal{X}} |fg|^r \ \mathrm{d}\mu \leq \frac{r}{p} \frac{1}{\|f\|_{p}^{p}} \int_{\mathcal{X}}|f|^{p} \ \mathrm{d}\mu + \frac{r}{q} \frac{1}{\|g\|_{q}^{q}} \int_{\mathcal{X}} |g|^{q} \ \mathrm{d}\mu.$$
Observe that $\int_{\mathcal{X}}|f|^{p} \ \mathrm{d}\mu = \|f\|_{p}^{p}$ and similarly for $g$. Hence: $$\frac{1}{\|f\|_{p}^r\|g\|_{q}^r}\int_{\mathcal{X}} |fg|^r \ \mathrm{d}\mu \leq \frac{r}{p}  + \frac{r}{q} = r\left(\frac{1}{p} + \frac{1}{q}\right) = 1.$$
Multiplying through and taking roots: $$\|fg\|_r^r \leq \|f\|_{p}^r\|g\|_{q}^r \implies \|fg\|_r \leq \|f\|_p\|g\|_q.$$ 
