Kernel, Image, Eigenvalues, Eigenvectors, And Norm Of Integral Operator On L²(-π, Π)

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Hey everyone! Let's dive into a fascinating problem in functional analysis involving an integral operator on the Hilbert space L²(-π, π). We're going to explore the operator's norm, kernel, image, eigenvalues, and corresponding eigenvectors. Buckle up, it's going to be an exciting journey!

Problem Statement: The Integral Operator T

We're given the Hilbert space H = L²(-π, π), which consists of all square-integrable functions on the interval [-π, π]. Our star player is the integral operator T, defined as follows:

Tf = cos²(x) ∫[-π, π] sin(2y)f(y) dy

where f(x) belongs to L²(-π, π). Our mission, should we choose to accept it, is to determine the following characteristics of T:

  1. Kernel (Ker T): The set of all functions f in L²(-π, π) that are mapped to the zero function by T.
  2. Image (Im T): The set of all functions in L²(-π, π) that can be obtained by applying T to some function in L²(-π, π).
  3. Eigenvalues: The scalars λ for which there exists a non-zero function f in L²(-π, π) such that Tf = λf.
  4. Eigenvectors: The non-zero functions f in L²(-π, π) that satisfy Tf = λf for some eigenvalue λ.
  5. Norm (||T||): A measure of the "size" or "strength" of the operator T.

Let's roll up our sleeves and begin this mathematical adventure!

1. Delving into the Kernel of T (Ker T)

So, first things first, let's figure out the kernel of T. Remember, the kernel, or null space, is the set of all functions f in L²(-π, π) that get mapped to zero when we apply T. In other words, we need to find all f such that Tf(x) = 0 for almost every x in [-π, π].

Looking at the definition of T, we have:

Tf(x) = cos²(x) ∫[-π, π] sin(2y)f(y) dy = 0

Now, this equation holds if either cos²(x) = 0 or the integral ∫[-π, π] sin(2y)f(y) dy = 0. Let's break it down:

  • Case 1: cos²(x) = 0 This happens when cos(x) = 0, which means x = ±π/2. However, we're dealing with functions in L²(-π, π), and a function being zero at a single point doesn't affect its equivalence class (since we consider functions equal if they are equal almost everywhere). So, this case doesn't give us much information about the kernel itself.

  • Case 2: ∫[-π, π] sin(2y)f(y) dy = 0 Aha! This is the key. The integral part is a scalar value. Let's call it C:

    C = ∫[-π, π] sin(2y)f(y) dy

    So, Tf(x) = cos²(x) * C. For Tf(x) to be zero, we need C = 0. This translates to:

    ∫[-π, π] sin(2y)f(y) dy = 0

    This integral equation tells us that f must be orthogonal to the function sin(2y) in L²(-π, π). In other words, the inner product of f and sin(2y) must be zero.

    Therefore, Ker T consists of all functions in L²(-π, π) that are orthogonal to sin(2y). We can express this as:

    Ker T = f ∈ L²(-π, π) <f, sin(2y)> = 0

    To further illustrate, we can say that Ker T is the orthogonal complement of the subspace spanned by sin(2y).

2. Unveiling the Image of T (Im T)

Next up, let's explore the image of T. The image, or range, of T is the set of all functions that can be obtained by applying T to some function in L²(-π, π). To figure this out, we need to analyze what the operator T actually does to a function.

Remember the definition:

Tf = cos²(x) ∫[-π, π] sin(2y)f(y) dy

Let's revisit our friend C:

C = ∫[-π, π] sin(2y)f(y) dy

As we saw before, C is just a scalar value. Now, notice something crucial: Tf(x) is simply cos²(x) multiplied by this scalar C. No matter what function f we start with, the output of T will always be a multiple of cos²(x).

This means that Im T is the subspace spanned by the function cos²(x). We can write this as:

Im T = g ∈ L²(-π, π) g(x) = C * cos²(x) for some scalar C

In simpler terms, Im T is a one-dimensional subspace of L²(-π, π) generated by the function cos²(x). This gives us a pretty clear picture of the functions that can result from applying T.

3. Hunting for Eigenvalues of T

Now for the exciting part: eigenvalues! An eigenvalue λ of T is a scalar such that Tf = λf for some non-zero function f (the eigenvector). This means that applying T to the eigenvector simply scales the eigenvector by a factor of λ.

Let's set up the eigenvalue equation:

Tf = λf(x)

Substituting the definition of T:

cos²(x) ∫[-π, π] sin(2y)f(y) dy = λf(x)

Again, let's use C:

C = ∫[-π, π] sin(2y)f(y) dy

So, we have:

C * cos²(x) = λf(x)

This equation is the key to finding the eigenvalues and eigenvectors. Let's analyze it:

  • Case 1: λ = 0 If λ = 0, then C * cos²(x) = 0. This means either C = 0 or cos²(x) = 0. We already know that cos²(x) = 0 only at isolated points, which doesn't affect the function in L²(-π, π) sense. So, we need C = 0. This leads us back to the kernel condition:

    ∫[-π, π] sin(2y)f(y) dy = 0

    This confirms that any function in Ker T is an eigenvector corresponding to the eigenvalue λ = 0.

  • Case 2: λ ≠ 0 If λ is not zero, we can rearrange the equation to solve for f(x):

    f(x) = (C / λ) * cos²(x)

    This tells us that the eigenvector f must be a multiple of cos²(x). In other words, the eigenvector lies in the image of T! Now, we need to find the specific value of λ that makes this equation consistent. Let's substitute this expression for f(x) back into the equation for C:

    C = ∫[-π, π] sin(2y) [(C / λ) * cos²(y)] dy

    If C = 0, we just get the trivial solution (f = 0), which we don't want. So, let's assume C ≠ 0 and divide both sides by C:

    1 = (1 / λ) ∫[-π, π] sin(2y) cos²(y) dy

    Now, we need to evaluate this integral. Remember the double angle identity: sin(2y) = 2sin(y)cos(y). So, we have:

    ∫[-π, π] sin(2y) cos²(y) dy = ∫[-π, π] 2sin(y)cos³(y) dy

    We can use a simple u-substitution: let u = cos(y), then du = -sin(y) dy. The integral becomes:

    -2 ∫ cos³(y) (-sin(y) dy) = -2 [cos⁴(y) / 4] evaluated from -π to π

    = -[cos⁴(π) - cos⁴(-π)] / 2 = -[1 - 1] / 2 = 0

    Wait a minute! The integral is zero! This means that we have a contradiction. Our assumption that λ ≠ 0 led us to a dead end. The only way this equation can hold is if C = 0, which brings us back to the case where λ = 0.

    Therefore, the only eigenvalue of T is λ = 0.

4. Identifying Eigenvectors of T

Since the only eigenvalue is λ = 0, the eigenvectors are simply the functions in the kernel of T. We already determined that:

Ker T = f ∈ L²(-π, π) <f, sin(2y)> = 0

So, the eigenvectors of T are all functions in L²(-π, π) that are orthogonal to sin(2y).

5. Determining the Norm of T (||T||)

Finally, let's tackle the norm of T. The norm of an operator is a measure of its "size" or "strength." For a bounded linear operator like T, the norm is defined as:

||T|| = sup||Tf|| / ||f|| f ∈ L²(-π, π), f ≠ 0

In other words, ||T|| is the largest factor by which T can stretch a vector. To find ||T||, we need to maximize the ratio ||Tf|| / ||f||.

Let's first find ||Tf||²:

||Tf||² = ∫[-π, π] |Tf(x)|² dx = ∫[-π, π] |cos²(x) ∫[-π, π] sin(2y)f(y) dy|² dx

Again, let C = ∫[-π, π] sin(2y)f(y) dy. Then:

||Tf||² = ∫[-π, π] |C * cos²(x)|² dx = |C|² ∫[-π, π] cos⁴(x) dx

Now, let's evaluate the integral ∫[-π, π] cos⁴(x) dx. We can use the power-reduction formula for cos²(x):

cos²(x) = (1 + cos(2x)) / 2

So,

cos⁴(x) = (cos²(x))² = [(1 + cos(2x)) / 2]² = (1 + 2cos(2x) + cos²(2x)) / 4

Now, apply the power-reduction formula again to cos²(2x):

cos²(2x) = (1 + cos(4x)) / 2

So,

cos⁴(x) = (1 + 2cos(2x) + (1 + cos(4x)) / 2) / 4 = (3 + 4cos(2x) + cos(4x)) / 8

Now we can integrate:

∫[-π, π] cos⁴(x) dx = (1/8) ∫[-π, π] (3 + 4cos(2x) + cos(4x)) dx

= (1/8) [3x + 2sin(2x) + (1/4)sin(4x)] evaluated from -π to π

= (1/8) [3π - (-3π)] = (6π) / 8 = (3π) / 4

So,

||Tf||² = |C|² (3π / 4)

Now, let's find |C|²:

|C|² = |∫[-π, π] sin(2y)f(y) dy|²

By the Cauchy-Schwarz inequality:

|∫[-π, π] sin(2y)f(y) dy|² ≤ ∫[-π, π] |sin(2y)|² dy ∫[-π, π] |f(y)|² dy

= ||sin(2y)||² ||f||²

Let's calculate ||sin(2y)||²:

||sin(2y)||² = ∫[-π, π] sin²(2y) dy = ∫[-π, π] (1 - cos(4y)) / 2 dy

= (1/2) [y - (1/4)sin(4y)] evaluated from -π to π

= (1/2) [π - (-π)] = π

So,

|C|² ≤ π ||f||²

Now we have:

||Tf||² = |C|² (3π / 4) ≤ (π ||f||²) (3π / 4) = (3π² / 4) ||f||²

Taking the square root:

||Tf|| ≤ (π√(3) / 2) ||f||

This gives us an upper bound for ||T||. To show that this bound is tight, we need to find a function f for which the equality holds. Let's try f(x) = sin(2x):

C = ∫[-π, π] sin(2y)sin(2y) dy = ∫[-π, π] sin²(2y) dy = π

||f||² = ∫[-π, π] sin²(2x) dx = π, so ||f|| = √π

Tf(x) = cos²(x) * C = π cos²(x)

||Tf||² = π² ∫[-π, π] cos⁴(x) dx = π² (3π / 4)

||Tf|| = π √(3π / 4) = (π√(3) / 2) √π = (π√(3) / 2) ||f||

So, we found a function for which the equality holds! Therefore, the norm of T is:

||T|| = (π√(3) / 2)

Conclusion: A Deep Dive into the Integral Operator T

Phew! What a journey! We've successfully determined the kernel, image, eigenvalues, eigenvectors, and norm of the integral operator T. Here's a quick recap:

  • Ker T: Functions orthogonal to sin(2y)
  • Im T: Subspace spanned by cos²(x)
  • Eigenvalues: Only λ = 0
  • Eigenvectors: Functions in Ker T
  • ||T||: (π√(3) / 2)

This problem showcases the beauty and power of functional analysis, allowing us to understand the behavior of operators on infinite-dimensional spaces. We've used concepts like orthogonality, inner products, the Cauchy-Schwarz inequality, and eigenvalue equations to unravel the properties of T. Keep exploring, guys, there's a whole universe of mathematical wonders out there!