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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.

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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

1 Answer 1

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 $\{0\}$.

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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

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