This is a homework question for my independent study Spectral Theory class, but my professor is pretty busy with other stuff so has not been able to help me much.

This space of complex vectors has $$ ||x|| = \sum_{i=1}^\infty |x_i|^2 < \infty, \quad \quad x=(x_1, x_2, \dots) $$ And the question asks to classify the entire spectrum for $L$.

So far I have found $L-\lambda$ to be 1-1 so it does not have a Discrete Spectrum.

I have found the range of $L-\lambda$ to not be dense in $H$ (this particular Hilbert Space) when $|\lambda| < 1 \implies |\lambda| < 1$ is in the Residual Spectrum.

Now I'm trying to find when (for what $\lambda$) the inverse operator on its range is continuous (bounded). If the inverse, $(L-\lambda)^{-1}$, is bounded for some $\lambda_b$ then $\lambda_b$ is in the Resolvent Spectrum. If for some $\lambda_u$, $(L-\lambda)^{-1}$ is unbounded, then $\lambda_u$ is in the Continuous Spectrum.

I have tried several approaches, but my current one is to use the Bounded Inverse Theorem to find when $(L-\lambda)^{-1}$ is bounded.

I have found that $L-\lambda$ is bounded as long as $|\lambda|<\infty$.

I need to find when $L-\lambda$ is bijective. I already know it is always 1-1 so I just need to find where it is onto. This is where I am stuck. The answer should be $|\lambda|=1$ is in the Continuous Spectrum and $|\lambda| > 1$ is in the Resolvent Spectrum.

I have: $$ (L-\lambda)x = (0-\lambda x_1, x_1 - \lambda x_2, x_2 - \lambda x_3, \dots) $$

Let $y \in H$ so $||y|| < \infty$. Then choose $x$ s.t. \begin{align*} -\lambda x_1 =& y_1 \\ x_1 - \lambda x_2 =& y_2 \\ x_2 - \lambda x_3 =& y_3 \\ \vdots& \\ \implies x_1 =& \frac{y_1}{-\lambda} \\ \implies x_2 =& \frac{y_2-x_1}{-\lambda} = \frac{y_2}{-\lambda} - \frac{y_1/-\lambda}{-\lambda} = \frac{y_2}{-\lambda} + \frac{y_1}{-\lambda^2} \\ \implies x_3 =& \frac{y_3-x_2}{-\lambda} = \frac{y_3}{-\lambda} - \frac{(y_2/-\lambda)+(y_1/-\lambda^2)}{-\lambda} = \frac{y_3}{-\lambda} + \frac{y_2}{-\lambda^2} + \frac{y_1}{-\lambda^3} \\ \vdots& \\ \implies x_n =& \sum_{i=1}^n \frac{y_i}{-\lambda^{n-i+1}} \\ \vdots& \\ \implies (L - \lambda)^{-1}y =& \left(\sum_{i=1}^1 \frac{y_i}{-\lambda^{1-i+1}}, \sum_{i=1}^2 \frac{y_i}{-\lambda^{2-i+1}}, \sum_{i=1}^3 \frac{y_i}{-\lambda^{3-i+1}}, \dots \right) \end{align*} so that \begin{align*} ||x|| =& \sum_{n=1}^\infty |x_n|^2 \\ =& \sum_{n=1}^\infty \left| \sum_{i=1}^n \frac{y_i}{-\lambda^{n-i+1}} \right|^2 \\ \leq& \sum_{n=1}^\infty \left( \sum_{i=1}^n \frac{|y_i|}{|\lambda^{n-i+1}|} \right)^2 \\ \leq& \sum_{n=1}^\infty \sum_{i=1}^n \frac{|y_i|^2}{|\lambda^{n-i+1}|^2} \\ =& \sum_{n=1}^\infty \sum_{i=1}^n \frac{|y_i|^2}{|\lambda|^{2(n-i+1)}} \end{align*}

If $|\lambda| = 1$ the above becomes \begin{align*} \sum_{n=1}^\infty \sum_{i=1}^n |y_i|^2 =& \lim_{n\to \infty} n|y_1|^2 + (n-1)|y_2|^2 + \dots + (n-(n-2))|y_{n-1}|^2 + (n-(n-1))|y_{n}|^2 \\ =& \lim_{n\to \infty} \left( n \sum_{i=1}^n |y_i|^2 - \sum_{i=1}^n (i-1)|y_i|^2 \right) \\ =& \lim_{n\to \infty} \left( n \sum_{i=1}^n |y_i|^2 + \sum_{i=1}^n |y_i|^2 \right) + \lim_{n\to \infty} \left( \sum_{i=1}^n -i |y_i|^2 \right) \\ =& ||y|| \lim_{n\to \infty} (n+1) + \lim_{n\to \infty} \left( \sum_{i=1}^n -i |y_i|^2 \right) \\ \end{align*}

I should get that $||x|| < \infty$ only when $|\lambda| \leq 1 \implies |\lambda| = 1$ is in the Continuous Spectrum and $|\lambda| > 1$ is in the Resolvent Spectrum. I can't seem to get this to work out.

I appreciate any help. Thanks!

This question is 7.1.2. a) from Keener's Principles of Applied Mathematics (revised edition).

  • 1
    $\begingroup$ I don't think $L - \lambda I$ is even subjective for $|\lambda| = 1$. Is $e_1 := (1, 0, \cdots)$ in the image? $\endgroup$
    – mathdoge
    Jun 15, 2020 at 23:24
  • 1
    $\begingroup$ Are you sure you wrote the definition of the norm correctly? Square root not lost? Show that the norm of the operator is equal to 1, thus its spectrum lies in a closed unit circle, therefore the operator $L-\lambda I$ is bounded outside it. $\endgroup$
    – thing
    Jun 16, 2020 at 2:36
  • $\begingroup$ @mathdoge: Thanks I think you are right. Is the Bounded Inverse Theorem an if and only if? So that showing $L-\lambda$ is not surjective will prove that its inverse is not bounded? $\endgroup$ Jun 16, 2020 at 16:56
  • $\begingroup$ @thing: Thanks for the comment - I wrote the norm as it appears in the problem. I see that $||Lx||/||x|| = 1$ so does that mean $$\frac{||(L-\lambda)x||}{||x||} \leq \frac{||Lx||}{||x||} + \frac{||-\lambda x||}{||x||} = 2$$ when $|\lambda|=1$? $\endgroup$ Jun 16, 2020 at 17:01
  • 1
    $\begingroup$ I am not sure, but if $L - \lambda I$ is not subjective, then it has no inverse. $\endgroup$
    – mathdoge
    Jun 16, 2020 at 17:07

1 Answer 1


From @mathdoge:

If $|\lambda| = 1$ and we choose as an example $v =(1, 0, 0, \dots)$, then clearly $||v|| = 1 < \infty$ so $v$ is in this vector space. For this $v$ and $|\lambda| = 1$, $||x||$ becomes \begin{align*} ||x|| =& \sum_{n=1}^\infty \left| \sum_{i=1}^n \frac{v_i}{-\lambda^{n-i+1}} \right|^2 \\ =& \sum_{n=1}^\infty 1 \\ =& \infty \end{align*} so $L - \lambda$ is not onto if $|\lambda|=1$. Then the Bounded Inverse Theorem implies that $|\lambda|=1$ is in the Continuous Spectrum.

Still need to show that $|\lambda|>1$ not in the Continuous Spectrum $\iff$ $|\lambda|>1$ in the Resolvent Spectrum.


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