- #1

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- Homework Statement
- $$\int_{0}^{\infty}e^{-t} f^2(t) dt$$

- Relevant Equations
- $$\int_{0}^{\infty}e^{-t} f(t)g(t) dt$$

I'm learning Linear Algebra by self and I began with Apsotol's Calculus Vol 2. Things were going fine but in exercise 1.13 there appeared too many questions requiring a strong knowledge of Real Analysis. Here is one of it (question no. 14)

I have only this much knowledge that

converges to ##I## means there exists an ##N## for each ##\epsilon## (no matter how small) such that whenever ##b \gt N## we have

$$

\lvert \int_{0}^{b} e^{-t} f^2(t) dt \rvert - I \lt \epsilon

$$

Now, I have this undeveloped idea in my head that somehow I have to replace ##f^2(t)= f(t) \times f(t)## by ##f(t) \times g(t)## as both ##f## and ##g## belong to ##V## and show that epsilon relation still exists. Now, I request for your benign guidance.

*Let ##V## be the set of all real functions ##f## continuous on ##[0, +\infty)## and such that the integral*

$$

\int_{0}^{\infty} e^{-t} f^2 (t) dt

$$

converges. Define ##\langle f, g \rangle = \int_{0}^{\infty} e^{-t} f(t) g(t) dt##.

Prove that the integral for ##\langle f, g\rangle## converges absolutely for each pair of functions ##f## and ##g## in ##V##.$$

\int_{0}^{\infty} e^{-t} f^2 (t) dt

$$

converges. Define ##\langle f, g \rangle = \int_{0}^{\infty} e^{-t} f(t) g(t) dt##.

Prove that the integral for ##\langle f, g\rangle## converges absolutely for each pair of functions ##f## and ##g## in ##V##.

I have only this much knowledge that

*$$*

\int_{0}^{\infty} e^{-t} f^2 (t) dt

$$\int_{0}^{\infty} e^{-t} f^2 (t) dt

$$

converges to ##I## means there exists an ##N## for each ##\epsilon## (no matter how small) such that whenever ##b \gt N## we have

$$

\lvert \int_{0}^{b} e^{-t} f^2(t) dt \rvert - I \lt \epsilon

$$

Now, I have this undeveloped idea in my head that somehow I have to replace ##f^2(t)= f(t) \times f(t)## by ##f(t) \times g(t)## as both ##f## and ##g## belong to ##V## and show that epsilon relation still exists. Now, I request for your benign guidance.