Show $\Vert f\Vert_{L_1}=\Vert f\Vert_{L_2}=p$ on $[0,1]$ implies $f\equiv p$ a. e.

Question

Let $f\colon [0,1]\rightarrow\mathbb{R}_0^+$ be an integrable function. Suppose there exists $p\in\mathbb{R}$ such that $$\int_{[0,1]}f(x)\,dx=p\enspace\text{ and }\enspace\int_{[0,1]}f(x)^2\,dx=p^2.$$ How does it follow that $f(x)= p$ for almost all $x\in[0,1]$?

Answer 1

GEdgar's hint is helpful. Compute \begin{equation} \int_{[0,1]} (f(x)-p)^2 dx = \int_{[0,1]} (f(x)^2-2pf(x) + p^2)dx = \int_{[0,1]} f(x)^2dx - 2p\int_{[0,1]} f(x) dx + p^2. \end{equation} If you substitute in your known values, the right side will vanish and you can make the desired conclusion.

Answer 2

Equality holds in the Cauchy-Schwarz inequality $\|f_1f_2\|_1\leq\|f_1\|_2\|f_2\|_2$ if and only if $f_1$ is a scalar multiple of $f_2$ almost everywhere. The hypotheses $$ \int_{[0,1]}f(x)dx=p\text{ and }\int_{[0,1]}f(x)^2 dx=p^2 $$ tell us that $\|1\cdot f\|_1=\|1\|_2\|f\|_2=\|f\|_2$, where $1$ is the constant function $1$ on $[0,1]$. Thus, $f$ must be constant almost everywhere, and thus (since $\|f\|_1=p$) must equal $p$.