Show an operator on L^2(0,\infty) is bounded

[economist13]

Homework Statement

Show that the operator on $$L^2(0,\infty)$$ defined by $$g \rightarrow f(x)= \int_{0}^{\infty} e^{-xy}g(y)dy$$ is bounded.

Homework Equations

Operator norm: $$||T|| = \sup_{||g||_{L^2}=1}||Tg||_{L^2}$$

The Attempt at a Solution

I tried to get a handle on $$f(x)= \int_{0}^{\infty} e^{-xy}g(y)dy$$, by first fixing x and applying Holder's inequality. I got that for every x, $$f(x) \leq \frac{1}{2x}$$ but this didn't really get me much since $$\int_{0}^{\infty}(\frac{1}{2x})^2dx$$ doesn't converge...

What I need is that $$\int_{0}^{\infty} | \int_{0}^{\infty}e^{-xy}g(y)dy|^2dx$$ is finite. any inequalities/tricks I'm not thinking of? I'm hoping I'm just not realizing what theorem/inequality I need to use...I really appreciate any help you all can offer

 

[mighty2000]

!

Dear fellow scientist,

The operator in question can indeed be shown to be bounded using the operator norm definition. Here's how:

First, we can rewrite the operator as Tg(x) = \int_{0}^{\infty} e^{-xy}g(y)dy = \int_{0}^{\infty} f(x,y)g(y)dy, where f(x,y) = e^{-xy}.

Then, we can use the Cauchy-Schwarz inequality to get:

|Tg(x)| = |\int_{0}^{\infty} f(x,y)g(y)dy| \leq \int_{0}^{\infty} |f(x,y)||g(y)|dy

= \int_{0}^{\infty} e^{-xy}|g(y)|dy.

Now, let's define h(x) = \int_{0}^{\infty} e^{-xy}|g(y)|dy.

Then, we have:

||Tg||_{L^2}^2 = \int_{0}^{\infty} |Tg(x)|^2dx \leq \int_{0}^{\infty} h(x)^2dx.

Now, we can use the fact that h(x) is a continuous function to show that it is bounded on the interval [0,\infty). This means that there exists a constant C such that h(x) \leq C for all x \geq 0.

Therefore, we have:

||Tg||_{L^2}^2 \leq \int_{0}^{\infty} h(x)^2dx \leq C^2 \int_{0}^{\infty} dx = C^2.

Thus, we have shown that ||T|| \leq C, which means that the operator T is bounded with a bound of C. This completes the proof.

I hope this helps! Let me know if you have any further questions or if you need any clarification. Keep up the great work!