I need to prove that:If $f, g ∈ C[0, 1]$ and $1 < p, q < ∞$ with $\frac{1}{p}+\frac{1}{q}=1$
Then $|f(x)g(x)| dx ≤(\int^1_0|f(x)|^p dx)^{\frac{1}{p}}(\int^1_0|g(x)|^q dx)^{\frac{1}{q}}$
In the proof above, I do not quite understand why the $p,q$ such that $\int^0_1|f(x)|^p dx=\int^0_1|g(x)|^q dx=1$ exists.
And as we already have $\frac{1}{p}+\frac{1}{q}=1$, which means that once we fix $p$, we fix $q$ as well. It seems that saying $\int^0_1|f(x)|^p dx=\int^0_1|g(x)|^q dx=1$ is to say $\int^0_1|f(x)|^p dx=1$ implies $\int^0_1|g(x)|^q dx=1$.
So I am confused. Could someone please explain why we are able to take $p,q$ such that $\frac{1}{p}+\frac{1}{q}=1$ and $\int^0_1|f(x)|^p dx=\int^0_1|g(x)|^q dx=1$?
Thanks in advance!
For your first question, suppose we take $\overline{f} = |f| / (\int_{0}^{1} |f|^p )^{1/p}$ where $p$ is any real number in $(1,+\infty$) (note that $\overline{f}$ is just the function $|f|$ times the real number $c=(\int_{0}^{1} |f|^p )^{-1}$), then:
$$ \int_{0}^{1} |\overline{f}|^{p} = \int_{0}^{1} \left(\frac{|f|}{\left(\int_{0}^{1} |f|^p \right)^{1/p}}\right)^{p} = \frac{1}{\int_{0}^{1} |f|^p}\int_{0}^{1} |f|^p=1 \qquad (1)$$
For your second question, if we assume $p,q \in (0,+\infty)$, once one among $p,q$ is fixed, say $p$, then you can just solve an equation to get the other:
$$ \frac{1}{q} = \frac{p-1}{p} \Longrightarrow q =\frac{p}{p-1} $$
Keeping this in mind, once you defined $\overline{f}$ and $(1)$ holds, take $q$ as above (so that $p^{-1} + q^{-1} = 1$ where $p$ is the one used in $(1)$) and repeat the same contruction letting $\overline{g} = |g| / (\int_{0}^{1} |g|^q)^{1/q} $. You will get:
$$ \int_{0}^{1} |\overline{g}|^q = 1 = \int_{0}^{1} |\overline{f}|^{p} \qquad (2) $$
In the end, if you want, just rename $\overline{f},\overline{g}$ as $f,g$ again.
Remark:
We need $p,q$ to be positive (in particular, strictly greater than $1$) because the key for the proof is the following inequality (Young's):
$$ ab \le \frac{a^p}{p} + \frac{a^q}{q} $$
Where $a,b $ are non-negative real numbers and $p,q$ are positive real numbers satisfying $1=\frac{1}{p} + \frac{1}{q}$. Note that the request $p,q > 0$ essentially forces $p,q \in (1,+\infty)$, indeed if for example $p=\frac{1}{2} > 0 $ we get $q= -1 < 0$ (and viceversa)!