In a functional analysis course I saw a claim that for $1\leq p_{1}\leq p_{2}\leq\infty$we have it that $L^{p_{2}}\subseteq L^{p_{1}}$
I have a few problems with the proof given, and I would appreciate some clarifications. The proof given is (my comments are in $[]$):
Acuatlly, we will prove that $$ id:\,(C([a,b]),||\ldotp||_{p_{2}})\to(C([a,b]),||\ldotp||_{p_{1}}) $$
is continuous.
Let $f\in C([a.b])$. fix $0<\alpha<1$ and define:
$$ A=\{x\in[a,b]\mid|f(x)|>1\}\,,\, B=\{x\in[a,b]\mid\alpha\leq|f(x)|<1\}\,,\, C=\{x\in[a,b]\mid|f(x)|<\alpha\} $$
Then: $$ \int_{a}^{b}|f|^{p_{1}}=\int_{A}|f|^{p_{1}}+\int_{B}|f|^{p_{1}}+\int_{C}|f|^{p_{1}} $$
$$ \leq\int_{A}|f|^{p_{2}}+\int_{B}1+\int_{C}\alpha^{p_{1}} $$
$$ \leq||f||_{p_{2}}^{p_{2}}+\mu(B)+(b-a)\alpha^{p_{1}} $$
[those $3$ line are denoted as $(1)$]
[I don't think that the next line follows from what's done already] $$ ||f||_{p_{1}}^{p_{1}}\geq\int_{B}|f|^{p_{2}}\geq\alpha^{p_{2}}\mu(B) $$
[denote the above as $(\star)$]
$$ \implies\mu(B)\leq\frac{||f||_{p_{2}}^{p_{2}}}{\alpha^{p_{2}}} $$
set this in $(1)$ and get $$ ||f||_{p_{1}}^{p_{1}}\leq||f||_{p_{2}}^{p_{2}}+\frac{||f||_{p_{2}}^{p_{2}}}{\alpha^{p_{2}}}+(b-a)\alpha^{p_{1}} $$
Set $$ \alpha=||f||_{p_{2}}^{\frac{1}{2}} $$
$$ \implies||f||_{p_{2}}^{p_{2}}+||f||_{p_{2}}^{\frac{p_{2}}{2}}+(b-a)\alpha^{p_{1}} $$
And we see the required continuity.
My question are:
How did we get the first inequality in $(\star)$ ?
Why can we set $\alpha=||f||_{p_{2}}^{\frac{1}{2}}$ ? i.e why $0<\alpha<1$ ?
Why what we want to prove follows from the continuity mentioned, and how did it follow from the last line ?
Why did we star with $f\in C([a,b])$ ? I thought the elements of $L^{p}$ just have integrability constraints on them, not continuity constraints (I guess that this relates to the last question)
Doesn't the last line just say that $\int|f|^{p_{1}}<\infty$ and that's what we want to prove ?
P.S I realize those are more questions here then a usual post, I will give $150$ point bounty to the accepted answer.