Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $k \in C\left( \left[ 0,1 \right] \times \left[ 0,1 \right],\mathbb{C} \right)$ is a given continuous complex function on $\left[ 0,1 \right] \times \left[ 0,1 \right]$. Let $B \in B\left( C\left( \left[ 0,1 \right] \right) \right)$ be bounded linear map from continuous complex functions on $\left[ 0,1 \right]$ to the same space, given by

$$\left( Bu \right)\left( s \right) = \int_0^s k\left( s,t \right)u\left( t \right) \; dt,$$

for $u \in C\left( \left[ 0,1 \right] \right),s \in \left[ 0,1 \right]$. Determine the spectral radius and spectrum of $B$.

My attempt:

$$\left\| Bu \right\|_\infty = \left\| \int_0^s k\left( s,t \right)u\left( t \right)dt \right\|_\infty = \max \limits_{s \in \left[ 0,1 \right]} \left| \int_0^s k\left( s,t \right)u\left( t \right) \; dt \right| \leqslant \max_{s \in \left[ 0,1 \right]} \int_0^s \left| k\left( s,t \right)u\left( t \right) \right|dt \leqslant \left\| u \right\|_\infty \max_{s \in \left[ 0,1 \right]} \int_0^s \left| k\left( s,t \right) \right| \; dt \Rightarrow \left\| B \right\|_\infty \leqslant \max_{s \in \left[ 0,1 \right]} \int_0^s \left| k\left( s,t \right) \right|\;dt $$

which implies that for spectral radius $\nu \left( B \right)$ of $B$, we have $\nu \left( B \right) \leqslant \max\limits_{s \in \left[ 0,1 \right]} \int_0^s \left| k\left( s,t \right) \right| \; dt $.

However, it seems to me that I am not any closer to the solution.

share|cite|improve this question
Yeah, function $k$ here doesn't have to be of the form $g\left( {s - t} \right)$ which would prove useful. However, that is not the case here :( – Alen Apr 9 '12 at 22:48
Also, ${L^2}$ is a Hilbert space, while, $\left( {C\left( {\left[ {0,1} \right],\mathbb{C}} \right),{{\left\| {} \right\|}_\infty }} \right)$ is only a Banach space. – Alen Apr 10 '12 at 0:36
up vote 1 down vote accepted

I haven't dotted all my i's here, so be gentle please...

I believe that $B$ is compact (using Arzela-Ascoli). The Fredholm Alternative means that you only need to look for a point spectrum. We know $0$ is in the spectrum, since $B$ is compact, so we need only look for solutions of $Bu = \lambda u$, with $\lambda \neq 0$.

Suppose a solution exists, then consider the equation $u = \frac{1}{\lambda} Bu$. Let $\overline{k} = \sup_{t,\tau \in [0,1]} | k(t,\tau)|$, then we have the estimate: $|u(t)| \leq \int_0^{t} \frac{\overline{k}}{\lambda}| u(\tau)| d\tau$. Now use the Gronwall-Bellman (or whoever your favorite discoverer is) inequality to conclude that $u(t) = 0$. Hence $B$ has no point spectrum with $\lambda \neq 0$.

The spectral radius is therefore $0$, and the spectrum is $\{0\}$.

share|cite|improve this answer
Wow, neat, except I've never heard of those names. But thank you for you answer, as soon as I'm finished checking it, I'll be back to upvote and/or ask aditional questions, if any. Thanks again – Alen Apr 10 '12 at 1:34
Rudin's "Functional Analysis" has all the relevant results except the integral inequality which you will find in a book on ODEs. – copper.hat Apr 10 '12 at 2:32
Sorry, got cut off by the 5 min. editing limit. Wikipedia has an entry on the integral inequality. Compact operators are also referred to as completely continuous. A useful reference is Kolmogorov & Fomin's "Introductory Real Analysis", Section 24 may be germane. The Fredholm Alternative basically means that all non-zero members of the spectrum are eigenvalues, so you do not have to worry about the continuous or residual elements of the spectrum. – copper.hat Apr 10 '12 at 2:42
Thanks, I found it all at the end of our textbook, we are currently in the middle, hence the confusion. This problem is completely inappropriate considering my current knowledge. I understand all your steps and there is no flaw in any of them, so thank you again, my hat is off to you. Thank you very much, again – Alen Apr 10 '12 at 2:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.