I stumbled upon the proof of the proposition below in my lecture notes, and I have some difficulties understanding one part of the proof.
The proposition is:
Suppose $ \mu (X) < \infty$. If $f \in \mathcal{L}^p (X, \mathcal{M}, \mu)$, then $\forall q \in [1, p)$, $f \in \mathcal{L}^q(X, \mathcal{M}, \mu)$ and we have: $$ ||f||_q \leq ||f||_p \mu(X)^{\frac{1}{q} - \frac{1}{p}} $$
The proof begins with the case when $p = \infty$, which is pretty straightforward for me. Then, the proof goes as follows:
If $P<\infty$ and $1 \leq q < p$, then we have: $$\int_{X}|f|^qd \mu = \int_{X}|f|^q \cdot 1 d\mu \leq \left( \int_{X}|f|^pd\mu \right)^{\frac{q}{p}}\mu(X)^{1 - \frac{q}{p}}$$
And this is where I get stuck. I don't see at all how we get the inequality. I suppose we use the Hölder's inequality somehow, but I can't figure out how exactly. Can someone show me how we get the last inequality?
You use Hölder's inequality in the following way:
\begin{align*} \int_X \vert f \vert^q \, d\mu = \Vert \vert f \vert^q \cdot \mathbf 1 \Vert _1 \leq \Vert \vert f \vert^q \Vert_{\frac p q} \Vert \mathbf 1 \Vert_{\frac{p}{p - q}} &= \bigg(\int_X (\vert f \vert^q)^{\frac p q} \, d\mu \bigg)^{\frac q p} \bigg(\int_X \mathbf 1^{\frac{p}{p - q}} \, d\mu \bigg)^{\frac{p - q}{p}} \\ &= \bigg(\int_X \vert f \vert^p \, d\mu \bigg)^{\frac q p} \bigg(\int_X \mathbf 1 \, d\mu \bigg)^{1 - \frac{q}{p}} \\ &= \bigg(\int_X \vert f \vert^p \, d\mu \bigg)^{\frac q p} \mu(X) ^{1 - \frac{q}{p}}. \end{align*} Notice esspiecally that $\frac p q + \frac{p}{p - q} = 1$ and $\mathbf 1 \in L^p(X)$ (where $\mathbf 1$ denotes the constant function with value $1$) for all $p \in [1, \infty)$ since your measure space is finite.