# Injective operator in L^2(0,1)

1. Jan 6, 2010

### Carl140

Hi!

Define an integral operator K: L^2 (0,1) -> L^2(0,1) by:

Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from t=0 to t=1].

Why is "obvious" that K is a one-to-one operator?

I know K is one to one if Kx(t) = 0 implies x(t) = 0 but I don't see why this is true. Can you please explain why?

2. Jan 6, 2010

### Landau

Please use Latex. Do you mean the following? (you wrote 'from t=0 to t=1' while s is your integration variable)

$$Kx(t)=\int_0^1 (1+ts)\exp(ts)x(s)\mbox{d}s$$

3. Jan 6, 2010

### Carl140

Yes, sorry about that. It is from s=0 to s=1.

4. Jan 12, 2010

### brahman

$$L^2(0,1)$$ is a Hilbert separable space with inner product
$$<u,v> = \int_0^1 uv dx$$​
and the set of Legendre polynomials is a Hilbert base of $$L^2(0,1)$$
$$P_n (x) = \frac{\sqrt{2n+1}}{ 2^{n+1/2} n!} \frac{d^n}{dx^n} \left[ (x^2 - 1)^n \right] \; \; , \; n = 1,2,3,...$$​
It mean
$$\forall v \in L^2(0,1) , v = \sum_{n=1}^{\infty} v_n P_n$$​
where $$v_n = <v, P_n>$$.

Now we assume $$x \in L^2 (0,1) \, , \, Kx = 0$$ and try to prove that $$x = 0$$.

First put $$k(t,s) =(1+ts)\exp(ts)$$, by applying the dominate convergence theorem, we have
$$\frac{d}{dt} \int _0^1 k(s,t) x(s) ds = \int _0^1 \frac{\partial}{\partial t}k(s,t) x(s) ds$$​
and further we have
$$\frac{d^n Kx}{dt^n} (0) = \left. \left( \frac{d^n}{dt^n} \int _0^1 k(s,t) x(s) ds \right) \right|_{t= 0}= \int _0^1 \left. \left( \frac{\partial^n}{\partial t^n}k(s,t) \right) \right|_{t=0} x(s) ds \;\;\; (1)$$​
by applying the induction principle we get
$$\frac{\partial^n}{\partial t^n}k(s,t) = [(n+1)s^n + ts^{n+1} ] \exp(ts) \; ,\; \forall n = 1,2,3,... \;\;\; (2)$$​
Combining (1)(2) and assumption Kx = 0, we get then
$$\int_{0}^{1} s^n x(s) = 0 \; , \; \forall n = 1,2,3, ... \;\;\; \;\;\; (*)$$​
From $$(*)$$ we have
$$x_n = <x, P_n> = 0 \; ,\; \forall n = 1,2,3,...$$​
finally
$$x = \sum_{n=1}^{\infty} x_n P_n= 0$$​