# Showing it is an orthogonal projector

Let $\phi:[0,1]\to\mathbb R$ contionuous and $A:L_2([0,1])\to L_2([0,1])$ defined by

$(Af)(x)=\phi(x)\int_{0}^{1}\phi(t)f(t)dt$

I already showed that $A=A^*$ and that $A$ is positive, but I would like to know two things:

1) When is A an orthogonal projector?

2) Is there a $\lambda\ge0$, such that $A^2=\lambda A$

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## 1 Answer

• For 1) When $\int |\phi(t)|^2dt=1$,
• For 2) Choose $\lambda=\int |\phi(t)|^2dt$.

In fact the main observation to do is that

\begin{align*} (AAf)(x) & =\phi(x)\int_0^1\phi(t)\phi(t)\int_0^1\phi(s)f(s)dsdt\\ & =\int_0^1|\phi(t)|^2dt \phi(x) \int_0^1\phi(s)f(s)ds\\ & =\int_0^1|\phi(t)|^2dt Af(x)\,.\end{align*}

Also, for any continuous function $\psi:[0,1]\to\mathbb R$, with $\psi(x)\neq0$ for some $x\in[0,1]$ the function $$\phi=\frac{\psi}{(\int_0^1|\psi(x)|^2dx)^{1/2}}$$ will give you a projector.

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1) $\phi(t)=1$ ? – Voyage Nov 28 '12 at 15:34
$\phi:t\mapsto 1$ is one of the possible $\phi$ giving an application $A$ which is a projector. – Sebastien B Nov 28 '12 at 15:44