# Prove that the inner product converges

• Hall
In summary, the problem is that the integrand in the Cauchy-Schwartz inequality is not a continuous function, so the proof does not work. However, if f is continuous and |f| is integrable on [0, ∞), then so is f.f

#### Hall

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)

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##.

I have only this much knowledge that
$$\int_{0}^{\infty} e^{-t} f^2 (t) dt$$

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.

You don't necessarily need a proof from first principles (involving ##\epsilon##), but you do need to understand the properties of limits and integrals.

One technique in this case is to let ##a > 0##, and look at:
$$\langle f, g \rangle_a = \int_0^a e^{-t} f(t)g(t) dt$$
Apostol shows that this is a well-defined inner product for the set of functions where the following integral exists:
$$\int_0^a e^{-t} f(t)^2 dt$$
Then use the Cauchy-Schwartz inequality for this inner product to prove the convergence of the integral in the case where ##a \rightarrow \infty##.

Using an ##\epsilon## argument would be useful to formalise this, but first you could sort out the steps in the proof.

FactChecker
but first you could sort out the steps in the proof.
All right, but how to begin?
We know
$$\int_{0}^{a} e^{-t} f^2(t) dt$$
exists. And we know
$$\int_{0}^{a} e^{-t} g^2(t) dt$$
exists.

Now, use Cauchy-Schwartz. (I've just looked at Apostol and he gives precisely the same hint as I did!)

FactChecker
$$\lvert \langle f, g \rangle \rvert \leq \langle f,f\rangle \langle g,g\rangle$$
$$\lvert \int_{0}^{a} e^{-t} f(t)g(t) dt\rvert \leq \int_{0}^{a} e^{-t} f^2(t) dt ~ \int_{0}^{a} e^{-t} g^2(t)dt$$
If the RHS is ##I_1\times I_2## then we have our LHS being always less than that specific number and hence it exists?

$$\lvert \langle f, g \rangle \rvert \leq \langle f,f\rangle \langle g,g\rangle$$
$$\lvert \int_{0}^{a} e^{-t} f(t)g(t) dt\rvert \leq \int_{0}^{a} e^{-t} f^2(t) dt ~ \int_{0}^{a} e^{-t} g^2(t)dt$$
If the RHS is ##I_1\times I_2## then we have our LHS being always less than that specific number and hence it exists?
If ##f, g## are continuous, then
$$\int_{0}^{a} e^{-t} f(t)g(t) dt$$exists, as any continuous function is integrable on a finite range.

But, clearly, that's not true for all all continuous functions on ##[0, \infty)##.

However, if ##f## is continuous and ##|f|## is integrable on ##[0, \infty)## then so is ##f##.

The trick here is to show that ##|e^{-t}f(t)g(t)|## is integrable on ##[0, \infty)##.

The trick here is to show that |e−tf(t)g(t)| is integrable on [0,∞).
It seems to me I cannot do that.

But I definitely can show that ##\int_{0}^{\infty} e^{-t} f(t) g(t) dt## is bounded.

It seems to me I cannot do that.
The trick is to look at ##|f|## and ##|g|##::
$$\big ( \int_0^a e^{-t}|f(t)|g(t)| dt \big ) ^2 \le \big ( \int_0^a e^{-t}|f(t)|^2 dt \big )\big ( \int_0^a e^{-t}|g(t)|^2 dt \big )$$
Now, the LHS is an increasing function of ##a##. And the RHS is bounded by the product of the two infinite integral of ##f^2## and ##g^2##.

That means that the LHS converges to some real number as ##a \rightarrow \infty##, hence ##|fg|## is integrable on ##[0, \infty)##. Hence ##fg## is integrable on ##[0, \infty)##.

Note also that:
$$\big | \int_0^{\infty} e^{-t}f(t)g(t) dt \big | ^2 \le \big ( \int_0^{\infty} e^{-t}|f(t)|g(t)| dt \big ) ^2\le \big ( \int_0^{\infty} e^{-t}|f(t)|^2 dt \big )\big ( \int_0^{\infty} e^{-t}|g(t)|^2 dt \big )$$$$= \big ( \int_0^{\infty} e^{-t}f(t)^2 dt \big )\big ( \int_0^{\infty} e^{-t}g(t)^2 dt \big )$$

The most important thing to learn in mathematics is to use the theorems. Once you know that you are dealing with an inner product space, use the theorems for inner product spaces. Do not go back to basics like epsilon, delta proofs unless you have to. Cauchy-Schwarz is a very important inequality.

And the RHS is bounded by the product of the two infinite integral of f2 and g2.

That means that the LHS converges to some real number as a→∞,
Well, we got that LHS is bounded and that surely tells us that the limit on LHS doesn't diverge, but we still have to rule out one possibility: it doesn't even oscillate. Only then we can conclude that the LHS converges.

For each $t \in [0,\infty)$, apply the AM-GM inequality to $|f(t)|$ and $|g(t)|$: $$|f(t)||g(t)| \leq \frac{(|f(t)| + |g(t)|)^2}{4}.$$ Hence $$|f(t)g(t)| \leq \frac{|f(t)|^2 + |g(t)|^2}{2}.$$ Now multiply by $e^{-t}$, integrate from $0$ to $R > 0$ and take the limit $R \to \infty$.

Well, we got that LHS is bounded and that surely tells us that the limit on LHS doesn't diverge, but we still have to rule out one possibility: it doesn't even oscillate. Only then we can conclude that the LHS converges.

If $h$ is non-negative then $\int_0^x h(t)\,dt$ is an increasing function of $x$.

Hall
Now multiply by e−t, integrate from 0 to R>0 and take the limit R→∞.
So, we would get
$$\int_{0}^{\infty} e^{-t} \big|f(t) g(t)\big| dt = L$$
Where ##L## is a real number.

Do we proceed like this
$$\int_{0}^{\infty} e^{-t} f(t) g(t) dt \leq \int_{0}^{\infty} e^{-t} \big|f(t) g(t)\big| dt$$
$$\int_{0}^{\infty} e^{-t} f(t) g(t) dt \leq L \\ int_{0}^{\infty} e^{-t} f(t) g(t) dt = M$$
Where M is a real number?
And conclude that ##\langle f,g\rangle## exists?

Well, we got that LHS is bounded and that surely tells us that the limit on LHS doesn't diverge, but we still have to rule out one possibility: it doesn't even oscillate. Only then we can conclude that the LHS converges.
It can't oscillate, as it's an increasing function of ##a##.

So, we would get
$$\int_{0}^{\infty} e^{-t} \big|f(t) g(t)\big| dt = L$$
Where ##L## is a real number.

Do we proceed like this
$$\int_{0}^{\infty} e^{-t} f(t) g(t) dt \leq \int_{0}^{\infty} e^{-t} \big|f(t) g(t)\big| dt$$
$$\int_{0}^{\infty} e^{-t} f(t) g(t) dt \leq L \\ int_{0}^{\infty} e^{-t} f(t) g(t) dt = M$$
Where M is a real number?
And conclude that ##\langle f,g\rangle## exists?
That's why I stated that:
However, if ##f## is continuous and ##|f|## is integrable on ##[0, \infty)## then so is ##f##.
The proof of that requires a bit of ##\epsilon## magic. The idea is that eventually $$\int_b^{\infty}|f(t)|dt$$ must be small, hence so is $$|\int_b^{\infty}f(t)dt|$$

That's why I stated that:
Now, I realize its importance. But how did that occur to you? Is it something we do generally in real analysis?
And I'm still think how Cauchy-Schwarz inequality could give us an inequality involving ##\big| f(t) g(t)\big|##? ( I know you included it in Note but I couldn't grasp it).

Now, I realize its importance. But how did that occur to you? Is it something we do generally in real analysis?
And I'm still think how Cauchy-Schwarz inequality could give us an inequality involving ##\big| f(t) g(t)\big|##? ( I know you included it in Note but I couldn't grasp it).
I realized it must be true, otherwise such a function would be a counterexample to the innner product. Also, while looking at this I realized that the functions must be continuous. That was missing from my first post, if you look closely.

It was a combination of trying to prove the result and realising what I needed to prove along the way. This is proof-finding technique.

Proofs generally are a different ball game from applied mathematics. You must be constantly thinking:

1) Is this true?
2) What happens if it's not true?
3) Can I find a countexample?
4) Can I prove that step?

That was also how I got the insight about using ##|f|## and ##|g|## instead of ##f## and ##g##, where the integral could oscillate as you suggested.

Hall