Spectrum of a Linear Operator question

Question

"Let $X = C([−1, 1])$ be the space of continuous real-valued functions on $[−1, 1]$ with sup norm, $\lvert\lvert f\rvert\rvert_{\infty}$

Define the linear operator $T : X → X $ by $$(Tf)(x) = x^2f(−x).$$

Prove that $T$ is bounded, and find its norm. Moreover, determine for which $λ ∈ R$ the operator $λI − T$ is bijective"

Where do I begin here?

Answer 1

$|(Tf)(x)| =x^2|f(-x)| \le 1 \cdot ||f||_{\infty},$ hence

$$||Tf||_{\infty} \le ||f||_{\infty},$$

This gives $||T|| \le 1$.

Can you proceed ?

Answer 2

To study the eigenvalues, we consider the equation $Tf(x)=\lambda f(x)$ or $x^2f(-x)=\lambda f(x)$. First, obviously $0$ is not an eigenvalue, because from the equation $x^2f(-x)=0$ follows $f(x)\equiv0$. Let be $\lambda\ne0$. Substituting in the equation $-x$ instead of $x$ we get $x^2f(x)=\lambda f(-x)$. From these two equations we get that $(x^4-\lambda^2)f(x)=0$. Since $f$ is continuous, we obtain $f(x)\equiv0$. Thus the operator $T-\lambda I$ has no eigenvalues and is injective.

To find the spectrum of $T$ we consider the equation $(T-\lambda I)f(x)=g(x)$ for continuous function $g(x)$ and substitute into it $-x$ instead of $x$. We'll get two equations: $x^2f(-x)-\lambda f(x)=g(x)$ and $x^2f(x)-\lambda f(-x)=g(-x)$. From them we get $(x^4-\lambda^2)f(x)=\lambda g(x)+x^2g(-x)$.

So, if $\lambda\not\in[-1,1]$, we get $f(x)=\frac{\lambda}{x^4-\lambda^2}g(x)+\frac{x^2}{x^4-\lambda^2}g(-x)$ -- continuous function for any continuous function $g$. With a such $\lambda$, the operator $T-\lambda I$ is surjective. If $\lambda\in[-1,1]$, the continuous function $g(x)\equiv1$ does not lie in the image of the $T-\lambda I$. Thus, the points $\lambda\in[-1,1]$ are points of the spectrum.

We additionally verify that these are points of the residual spectrum. Let be $\lambda\in[0,1]$, then we consider the equation $(x^4-\lambda^2)f(x)=\lambda g(x)+x^2g(-x)$ and substitute into it $x^4=\lambda^2$ (or $x^2=\lambda$, or $x=\pm\sqrt\lambda$). We get $\lambda(g(\sqrt\lambda)+g(-\sqrt\lambda))=0$. Let be a $\lambda\ne0$, then image of $T-\lambda I$ lies in the set of continuous functions for which $g(\sqrt\lambda)+g(-\sqrt\lambda)=0$. We denote this set $M$ and suppose that $M$ is dense in $C[-1,1]$ that is, for each $\varepsilon>0$ for each $f\in C[-1,1]$ exists $g\in M$ such that $\|f-g\|<\varepsilon$ or $\forall x\in[-1,1]$ $|f(x)-g(x)|<\varepsilon$. In particular, $|f(\sqrt\lambda)-g(\sqrt\lambda)|<\varepsilon$ and $|f(-\sqrt\lambda)-g(-\sqrt\lambda)|<\varepsilon$. From here and from the triangle inequality $|f(\sqrt\lambda)+f(-\sqrt\lambda)|<2\varepsilon$. Due to randomness epsilon $f(\sqrt\lambda)+f(-\sqrt\lambda)=0$, and if $f$ is an even function, then we obtain a contradiction.

In the case $\lambda=0$ the original equation gives $x^2f(-x)=g(x)$, so image of operator lies in the set of continuous functions for which $g(0)=0$. This set is likewise not dense in $C[-1,1]$. The case $\lambda\in[-1,0]$ is treated similarly.