Prove that the inner product converges

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.
  • #1
Hall
351
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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)

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.
 
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  • #2
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.
 
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  • #3
PeroK said:
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.
 
  • #4
Now, use Cauchy-Schwartz. (I've just looked at Apostol and he gives precisely the same hint as I did!)
 
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  • #5
$$
\lvert \langle f, g \rangle \rvert \leq \langle f,f\rangle \langle g,g\rangle
$$
$$
\lvert \int_{0}^{a} e^{-t} f(t)g(t) dt\rvert \leq \int_{0}^{a} e^{-t} f^2(t) dt ~ \int_{0}^{a} e^{-t} g^2(t)dt$$
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?
 
  • #6
Hall said:
$$
\lvert \langle f, g \rangle \rvert \leq \langle f,f\rangle \langle g,g\rangle
$$
$$
\lvert \int_{0}^{a} e^{-t} f(t)g(t) dt\rvert \leq \int_{0}^{a} e^{-t} f^2(t) dt ~ \int_{0}^{a} e^{-t} g^2(t)dt$$
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)##.
 
  • #7
PeroK said:
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.
 
  • #8
Hall said:
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:
$$\big | \int_0^{\infty} e^{-t}f(t)g(t) dt \big | ^2 \le \big ( \int_0^{\infty} e^{-t}|f(t)|g(t)| dt \big ) ^2\le \big ( \int_0^{\infty} e^{-t}|f(t)|^2 dt \big )\big ( \int_0^{\infty} e^{-t}|g(t)|^2 dt \big )$$$$ = \big ( \int_0^{\infty} e^{-t}f(t)^2 dt \big )\big ( \int_0^{\infty} e^{-t}g(t)^2 dt \big )$$
 
  • #9
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.
 
  • #10
PeroK said:
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.
 
  • #11
For each [itex]t \in [0,\infty)[/itex], apply the AM-GM inequality to [itex]|f(t)|[/itex] and [itex]|g(t)|[/itex]: [tex]
|f(t)||g(t)| \leq \frac{(|f(t)| + |g(t)|)^2}{4}.[/tex] Hence [tex]
|f(t)g(t)| \leq \frac{|f(t)|^2 + |g(t)|^2}{2}.[/tex] Now multiply by [itex]e^{-t}[/itex], integrate from [itex]0[/itex] to [itex]R > 0[/itex] and take the limit [itex]R \to \infty[/itex].

Hall said:
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 [itex]h[/itex] is non-negative then [itex]\int_0^x h(t)\,dt[/itex] is an increasing function of [itex]x[/itex].
 
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  • #12
pasmith said:
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?
 
  • #13
Hall said:
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##.
 
  • #14
Hall said:
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:
PeroK said:
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 $$|\int_b^{\infty}f(t)dt|$$
 
  • #15
PeroK said:
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).
 
  • #16
Hall said:
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.
 
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FAQ: Prove that the inner product converges

1. What is an inner product?

An inner product is a mathematical operation that takes two vectors and produces a scalar value. It is often used in linear algebra and functional analysis to measure the angle between two vectors, the length of a vector, and to define orthogonality.

2. How do you prove that an inner product converges?

The convergence of an inner product can be proven by showing that the limit of the inner product as the number of terms approaches infinity exists and is finite. This can be done using various convergence tests, such as the Cauchy-Schwarz inequality or the triangle inequality.

3. What is the significance of proving the convergence of an inner product?

Proving the convergence of an inner product is important in many areas of mathematics, including optimization, functional analysis, and numerical analysis. It allows us to determine if a sequence of vectors or functions is approaching a specific value or if it is diverging.

4. Can you give an example of proving the convergence of an inner product?

Sure, for example, if we have the sequence of vectors {v1, v2, v3, ...} and we want to prove that the inner product of this sequence converges, we can show that the limit of the inner product of vi and vi+1 as i approaches infinity exists and is finite. This can be done using the Cauchy-Schwarz inequality, which states that the absolute value of the inner product of two vectors is less than or equal to the product of their lengths.

5. Are there any applications of proving the convergence of an inner product?

Yes, there are many applications of proving the convergence of an inner product. One example is in optimization, where the convergence of the inner product can be used to prove the convergence of an optimization algorithm. It is also used in numerical analysis to determine the accuracy and stability of numerical methods.

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