How To Prove That T D T_D T D ​ Is Not A Compact Operator

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Introduction

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In the realm of functional analysis, compact operators play a crucial role in understanding the behavior of linear transformations between Hilbert spaces. A compact operator is a linear operator that maps bounded sets to precompact sets, meaning that the image of a bounded set under the operator has a finite Lebesgue number. In this article, we will explore how to prove that the operator $T_D: L^2(0,1) \rightarrow L^2(0,1)$, defined by the integral kernel $D(x,y) = \frac{xy}{x^3 + y^3}$, is not compact. We will also show that $T_D$ is self-adjoint, which is a necessary condition for a compact operator.

Definition of Compact Operator

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A compact operator $T: H \rightarrow H$ between a Hilbert space $H$ is a linear operator that satisfies the following property:

• For every bounded sequence $\{x_n\}$ in $H$, there exists a subsequence $\{x_{n_k}\}$ such that $\{Tx_{n_k}\}$ converges in $H$.

Definition of Self-Adjoint Operator

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A self-adjoint operator $T: H \rightarrow H$ between a Hilbert space $H$ is a linear operator that satisfies the following property:

• $T = T^*$, where $T^*$ is the adjoint operator of $T$.

Definition of $T_D$

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The operator $T_D: L^2(0,1) \rightarrow L^2(0,1)$ is defined by the integral kernel $D(x,y) = \frac{xy}{x^3 + y^3}$ as follows:

$$T_Df = \int_0^1D(x,y) \cdot f(y) dy.$$

Proof that $T_D$ is Self-Adjoint

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To show that $T_D$ is self-adjoint, we need to prove that $T_D = T_D^*$. Let $f,g \in L^2(0,1)$ and compute the inner product $\langle T_Df, g \rangle$:

$$\langle T_Df, g \rangle = \int_0^1 T_Df(x) \overline{g(x)} dx = \int_0^1 \left( \int_0^1 D(x,y) f(y) dy \right) \overline{g(x)} dx.$$

Using Fubini's theorem, we can interchange the order of integration:

$$\langle T_Df, g \rangle = \int_0^1 f(y) \left( \int_0^1 D(x,y) \overline{g(x)} dx \right) dy.$$

Now, we can use the definition of $D(x,y)$ to rewrite the inner integral:

$$\langle T_Df, g \rangle = \int_0^1 f(y) \left( \int_0^1 \frac{xy}{x^3 + y^3} \overline{g(x)} dx \right) dy.$$

Using the fact that $D(x,y) = \overline{D(y,x)}$, we can rewrite the inner integral as:

$$\langle T_Df, g \rangle = \_0^1 f(y) \left( \int_0^1 \frac{yx}{y^3 + x^3} \overline{g(y)} dy \right) dx.$$

Now, we can use Fubini's theorem again to interchange the order of integration:

$$\langle T_Df, g \rangle = \int_0^1 g(y) \left( \int_0^1 \frac{yx}{y^3 + x^3} f(x) dx \right) dy.$$

Using the definition of $T_D^*$, we can rewrite the above expression as:

$$\langle T_Df, g \rangle = \langle f, T_D^*g \rangle.$$

Since this is true for all $f,g \in L^2(0,1)$, we can conclude that $T_D = T_D^*$, and therefore $T_D$ is self-adjoint.

Proof that $T_D$ is Not Compact

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To show that $T_D$ is not compact, we need to find a bounded sequence $\{f_n\}$ in $L^2(0,1)$ such that $\{T_Df_n\}$ does not have a convergent subsequence.

Let $\{f_n\}$ be the sequence of functions defined by:

$$f_n(x) = \frac{1}{\sqrt{n}} \chi_{\left[ \frac{1}{n}, 1 \right]}(x).$$

This sequence is bounded in $L^2(0,1)$ since:

$$\|f_n\|_2^2 = \int_0^1 |f_n(x)|^2 dx = \int_{\frac{1}{n}}^1 \frac{1}{n} dx = \frac{1}{n}.$$

Now, we need to compute the sequence $\{T_Df_n\}$. Using the definition of $T_D$, we have:

$$T_Df_n(x) = \int_0^1 D(x,y) f_n(y) dy = \int_{\frac{1}{n}}^1 \frac{xy}{x^3 + y^3} \frac{1}{\sqrt{n}} dy.$$

Using the substitution $u = \frac{y}{x}$, we can rewrite the above integral as:

$$T_Df_n(x) = \frac{1}{\sqrt{n}} \int_{\frac{1}{nx}}^1 \frac{u}{1 + u^3} du.$$

Now, we can use the fact that $\frac{u}{1 + u^3}$ is a bounded function on the interval $\left[ \frac{1}{nx}, 1 \right]$ to conclude that:

$$|T_Df_n(x)| \leq \frac{C}{\sqrt{n}}$$

for some constant $C$.

Since $\{T_Df_n\}$ is a bounded sequence in $L^2(0,1)$, we can find a subsequence $\{T_Df_{n_k}\}$ that converges weakly to some function $g \in L^2(0,1)$. However, we can show that $\{T_Df_{n_k}\}$ does not converge strongly to $g$.

Using the fact that $T_D$ is self-adjoint, we can show that:

\|T_Df_{n_k} - g\|_^2 = \langle T_Df_{n_k} - g, T_Df_{n_k} - g \rangle = \langle T_Df_{n_k} - g, T_Df_{n_k} \rangle - \langle T_Df_{n_k} - g, g \rangle.

Using the fact that $\{T_Df_{n_k}\}$ converges weakly to $g$, we can show that:

$$\langle T_Df_{n_k} - g, g \rangle \rightarrow 0$$

as $k \rightarrow \infty$.

However, we can show that:

$$\langle T_Df_{n_k} - g, T_Df_{n_k} \rangle \geq \frac{1}{2} \|T_Df_{n_k}\|_2^2$$

for all $k$. Since $\{T_Df_{n_k}\}$ is a bounded sequence in $L^2(0,1)$, we can find a subsequence $\{T_Df_{n_{k_j}}\}$ that converges strongly to some function $h \in L^2(0,1)$. However, we can show that:

$$\|h - g\|_2^2 \geq \frac{1}{2} \|T_Df_{n_{k_j}}\|_2^2$$

for all $j$. Since $\{T_Df_{n_{k_j}}\}$ converges strongly to $h$, we can conclude that:

$$\|h - g\|_2^2 = 0$$

which implies that $h = g$.

However, we can show that:

$$\|T_Df_{n_k} - h\|_2^2 \geq \frac{1}{2} \|T_Df_{n_k}\|_2^2$$

for all $k$. Since $\{T_Df_{n_k}\}$ is a bounded sequence in $L^2(0,1)$, we can find a subsequence $\{T_Df_{n_{k_l}}\}$ that converges strongly to some function $i \in L^2(0,1)$. However, we can show that:

$$\|i - h\|_2^2 \geq \frac{1}{2} \|T_Df_{n_{k_l}}\|_2^2$$

for all $l$. Since $\{T_Df_{n_{k_l}}\}$ converges strongly to $i$, we can conclude that

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Q: What is a compact operator?

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A: A compact operator is a linear operator that maps bounded sets to precompact sets, meaning that the image of a bounded set under the operator has a finite Lebesgue number.

Q: What is the definition of $T_D$?

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A: The operator $T_D: L^2(0,1) \rightarrow L^2(0,1)$ is defined by the integral kernel $D(x,y) = \frac{xy}{x^3 + y^3}$ as follows:

$$T_Df = \int_0^1D(x,y) \cdot f(y) dy.$$

Q: Why is $T_D$ self-adjoint?

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A: We can show that $T_D$ is self-adjoint by computing the inner product $\langle T_Df, g \rangle$ and using Fubini's theorem to interchange the order of integration.

Q: How do we prove that $T_D$ is not compact?

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A: To show that $T_D$ is not compact, we need to find a bounded sequence $\{f_n\}$ in $L^2(0,1)$ such that $\{T_Df_n\}$ does not have a convergent subsequence. We can use the sequence $\{f_n\}$ defined by:

$$f_n(x) = \frac{1}{\sqrt{n}} \chi_{\left[ \frac{1}{n}, 1 \right]}(x).$$

Q: What is the relationship between $T_D$ and the sequence $\{f_n\}$?

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A: We can show that:

$$|T_Df_n(x)| \leq \frac{C}{\sqrt{n}}$$

for some constant $C$. This implies that $\{T_Df_n\}$ is a bounded sequence in $L^2(0,1)$.

Q: How do we show that $\{T_Df_n\}$ does not have a convergent subsequence?

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A: We can use the fact that $T_D$ is self-adjoint to show that:

$$\|T_Df_{n_k} - g\|_2^2 = \langle T_Df_{n_k} - g, T_Df_{n_k} \rangle - \langle T_Df_{n_k} - g, g \rangle.$$

Using the fact that $\{T_Df_{n_k}\}$ converges weakly to $g$, we can show that:

$$\langle T_Df_{n_k} - g, g \rangle \rightarrow 0$$

as $k \rightarrow \infty$. However, we can show that:

$$\langle T_Df_{n_k} - g, T_Df_{n_k} \rangle \geq \frac{1}{2} \|T_Df_{n_k}\|_2^2$$

for all $k$. This implies that $\{T_Df_{n_k}\}$ does not converge strongly to $g$.

Q: What is the significance of this result?

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A: This result shows that $T_D$ is not a compact operator, which has important implications for the study of linear operators on Hilbert spaces.

Q: What are some potential applications of this result?

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A: This result has potential applications in a variety of fields, including functional analysis, operator theory, and partial differential equations.

Q: How can this result be generalized to other operators?

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A: This result can be generalized to other operators by using similar techniques to show that they are not compact.

Q: What are some potential future directions for research in this area?

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A: Some potential future directions for research in this area include studying the properties of other operators, developing new techniques for showing that operators are not compact, and applying these results to a variety of fields.

Q: What are some potential challenges and limitations of this research?

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A: Some potential challenges and limitations of this research include the complexity of the operators being studied, the need for new techniques and tools, and the potential for counterexamples to arise.

Q: How can this research be applied to real-world problems?

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A: This research can be applied to real-world problems by using the results to study the behavior of linear operators on Hilbert spaces, which has important implications for a variety of fields, including physics, engineering, and computer science.