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Suppose I have an operator on $L^2([0,1])$ defined by $f \longmapsto xf$. The eigenvalues occur at when $xf = \lambda x \implies \lambda = x$? Is this enough to conclude that $\sigma(T) = [0,1]$?

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    "Then clearly the eigenvalues are $\lambda=x$". What do you mean? – 2017-02-21
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    Is $x$ fixed, or do you mean the operator which sends $f$ to the map $x\mapsto xf(x)$? In the first case the spectrum is just $\{x\}$. – 2017-02-21
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    I assume by $xf$ you mean the function that takes $x\in [0,1]$ to $xf(x)$. Call the operator $A$. An eigenvalue $\lambda$ would be such that, for some $f$, $Af=\lambda f$, where $\lambda$ is fixed; but $Af=xf\ne\lambda f$ for all $x\ne\lambda$ where $f(x)\ne0$, i.e., for almost all $x$. – 2017-02-21
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    @ForgotALot Thanks, but how do I calculate the spectrum? – 2017-02-21
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    https://math.stackexchange.com/questions/975072/spectrum-of-multiplication-operator-by-the-independent-variable-in-l2?rq=1 – 2017-05-25

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No! There are no eigenvalues, but the spectrum is indeed $[0,1]$.

EDIT To see that this is the spectrum, calling your operator $T$, note that if $\lambda \notin [0,1]$, $(T - \lambda I)^{-1}$ is multiplication by $1/(x-\lambda)$ (which is bounded since that is a bounded function on $[0,1]$), while if $\lambda \in [0,1]$ there is no $f \in L^2[0,1]$ for which $(x - \lambda) f(x) = 1$.

However, there are no eigenvalues: an eigenvalue for $\lambda$ would be a nonzero $f \in L^2[0,1]$ such that $T f - \lambda f = 0$, i.e. $(x - \lambda) f(x) = 0$ a.e. in $[0,1]$. But that makes $f(x) = 0$ a.e. in $[0,1]$.

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    Could you elaborate? – 2017-02-21