Proving No Eigenvalues Exist for an Operator on a Continuous Function Space

[member 428835]

Homework Statement

Show no eigenvalues exist for the operator $Af(x) = xf(x)$ where $A:C[0,1]\to C[0,1]$.

Relevant Equations

Nothing comes to mind.

Eigenvalues $\lambda$ for some operator $A$ satisfy $A f(x) = \lambda f(x)$. Then

$$
Af(x) = \lambda f(x) \implies\\
xf(x) = \lambda f(x)\implies\\
(\lambda-x)f(x) = 0.$$

How do I then show that no eigenvalues exist? Seems obvious one doesn't exists since $\lambda-x \neq 0$ for all $x\in [0,1]$.

 

[MathematicalPhysicist]

Well the argument is subtle.

Since $x$ ranges over every point in $[0,1]$ and the supposed eigenvalue is some constant you cannot have $\lambda=x$, since the function $x$ is not constant.
And obviously $f(x)\ne 0$, otherwise it wouldn't be an eigenvalues' problem.

 

[member 587159]

You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.

 

[member 428835]

You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.

This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.

 

[member 587159]

This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.

It also works for $\lambda = 0$. Then your equation leads to $xf(x) = 0 \forall x$ and for $x \neq 0$ we then have $f(x) = 0$. By continuity again also $f(0) = 0$, so $f=0$ everywhere.

So, to answer your question concretely, I don't think doing part (b) separately is necessary.

Note that eigenvalue $\lambda = 0$ can only happen when $A$ is non-injective. This map is injective however (continuity is key again here). Maybe that's what (b) wants you to write?