Injective operator in L^2(0,1)

[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?

 

[Landau]

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

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

 

[Carl140]

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

 

[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

$$x = \sum_{n=1}^{\infty} x_n P_n= 0$$​