$L^p [0,1] = \{f: [0,1] \to \mathbb R$ ; $f$ measurable such that $\int_{0}^{1}|f|^p dx < + \infty\}$ and, $\parallel f \parallel_{p} = \int_{0}^{1} (|f|^{p})^{\frac{1}{p}}$. Let $F$ be a subspace of $L^{p}([0,1])$ such that $F = \{f \in L^{p}([0,1])$ ; $f(x)=0$ if $x > \frac{1}{2}$ }.
Let $T : L^{p}([0,1]) \to F$ defined by:
$T(f)(x) = f(2x) C_p$ if $x \leq \frac{1}{2}$ and $T(f)(x) = 0$ if $x > \frac{1}{2}$.
I need to find the constant $C_p$.
I proceed by,
We know that $\parallel Tf \parallel = \parallel f \parallel$. Then, $\parallel Tf \parallel$$^p$$_p$ $= \parallel f \parallel$$^p$$_p$ $ = \int_{0}^{1} |f|^{p} dx$.
And $\parallel Tf \parallel$$^p$$_p$ $ = \int_{0}^{\frac{1}{2}}$$|C_p.f(2x)|^p dx$.
Then $\int_{0}^{\frac{1}{2}}$$|C_p.f(2x)|^p dx$ $= \int_{0}^{1} |f|^{p} dx$.
Is it right? How could I find the constant $C_p$?
Put $y=2x$ to get $\int_0^{1}|f(y)|^{p} \frac 1 2 dy \, C_p^{p}=\int_0^{1}|f(x)|^{p}\, dx$. This gives $C_p^{p}=2$ or $C_p=2^{1/p}$.