Dimension of null space of an operator $T$

Question

Let, $K(x,y)$ be a kernal in $[0,1]\times [0,1]$ defined as $K(x,y)=\sin(2\pi x)\sin(2\pi y)$. Consider the integral operator $$T(u)(x)=\int_0^1 u(y)K(x,y)\,dy$$ where, $u\in C[0,1]$. Which of the following assertions on $T$ are true ?

(A) The null space of $T$ is infinite dimensional.

(B) $T$ has no negaive eigen value.

(C) $T$ has an eigen value greater than $3/4$.

From the given operation, null spce of $T$ is given by $$\{u\in C[0,1]: \int_0^1 u(y)\sin(2\pi y)dy=0\}.$$ From this, what can we say about the dimension?

Again if $\lambda$ is eigen value and $f(x)$ is eigen vector then, $Tf(x)=\lambda f(x)$. From this relation, I am not getting anything! Any help please.

Answer 1

Let's rewrite $T(u)(x)$ as $$ T(u)(x)=\sin(2\pi x)\int_0^1u(y)\sin(2\pi y)\,dy. \tag{1} $$ One can easily show that the integral in $(1)$ vanishes if $u(x)=\cos(2n\pi x)\,(n\in\mathbb{N})$ or $u(x)=\sin(2n\pi x)\,(n\in\mathbb{N}_{\geq 2})$. Since these functions are linearly independent, and they span an infinite dimensional subspace of $C[0,1]$, it follows that (A) is true. It also follows from $(1)$ that, if $T$ has a nonzero eigenvalue, then its associated eigenfunction must be $u(x)=C\sin(2\pi x)$. Inserting this function in $(1)$ we find $$ T(C\sin(2\pi x))=\sin(2\pi x)\int_0^1C\sin^2(2\pi y)\,dy=\frac{1}{2}C\sin(2\pi x). \tag{2} $$ We conclude, therefore, that $(B)$ is true and $(C)$ is false.

Answer 2

Let us try the following substitution:

$$t:=y-1/2\implies dy=dt\implies T(u)x:=\int_{-1/2}^{1/2}u\left(t+\frac12\right) K\left(x,t+\frac12\right)dt$$

and we thus need to check what functions $\;u\;$ fulfill that

$$-\int_{-1/2}^{1/2}u\left(t+\frac12\right)\sin2\pi t\,dt=0$$

Observe that any polynomial of the form $\;u(x)=x^{2n}\;$ will be in the kernel of $\;T\;$ , since

$$\;\left(t+\frac12\right)^{2n}=\left(-t-\frac12\right)^{2n}\;$$

and thus the above integral is one of an odd function on a symmetric interval about zero and thus it equals zero.

Conclude that $\;\dim\ker T=\infty\;$