I will be putting a bounty on this problem as soon as it lets me. For those who want to understand where the problem came from I encourage reading the edits, as I cut out several failed attempts and no longer relevant definitions from the problem statement to avoid clutter.

Consider the integral operator $K: C([0,1])\to C([0,1])$ $$Kf(x) = \int_0^1k(x,y)f(y)dy$$ where $k(x,y) = x^2+2xy+y^2$.

Show that the image $I:=K(\overline{B}(0,1))$ of the closed unit ball in $C([0,1])$ using the supremum (maximum) norm $\left\Vert \cdot \right\Vert_\infty$ is closed.

Any of the following will be rewarded the bounty:

  1. Proof that $I$ is sequentially closed by showing if $Kf_n(x) \to F(x) \in C([0,1])$ then there is a formula for $f \in C([0,1])$ with $\left\Vert f \right\Vert_\infty\leq1$ such that $Kf(x) = F(x)$.
  2. Proof that $I$ is sequentially closed by showing if $Kf_n(x) \to F(x) \in C([0,1])$ then there there exists $f\in C([0,1])$ with $\left\Vert f \right\Vert_\infty\leq1$ such that $Kf(x) = F(x)$, i.e. a nonconstructive proof (perhaps one could apply the Baire Category Theorem?).
  3. Proof that $I$ is closed by showing that $C([0,1])\setminus I$ is open.
  4. Proof that $I$ is compact.
  5. Proof that $I$ is sequentially compact.
  6. Proof that $I$ is closed by some more clever method I haven't thought of.

Things I have proven which may or may not be useful to help you help me solve this:

  1. $C([0,1])$ with the maximum norm is a Banach space.
  2. If $f\in C([0,1])$ then $Kf \in C([0,1])$.
  3. If $Kf_n(x) \to F(x) \in C([0,1])$ then the convergence is uniform.
  4. If $Kf_n(x) \to F(x) \in C([0,1])$ this does not necessarily imply $f_n$ has a convergent subsequence (counterexample $f_ n(x)=\sin(nx)$).
  5. $K(\overline{B}(0,1))$ is bounded by $\overline{B}(0,7/3)$, hence $K$ is a continuous linear operator.
  6. If $Kf_n(x) \to F(x) \in C([0,1])$ then $F(x) = a+bx+cx^2$ for some $a\in[-1/3,1/3],b\in[-1,1],c\in[-1,1]$.
  7. If $Kf_n(x) \to F(x)=a+bx+cx^2$ then there is a function $f\in C([0,1])$ such that $Kf(x) = F(x)$, but it does not necessarily satisfy $\left\Vert f \right\Vert_\infty \leq 1$. The function is given by $$f(x)=(30a -18b+9c)+(-180a+96b-36c)x+(180a-90b+30c)x^2$$ and an example of when it fails then norm condition is if $$f_n(x)=e^{-x^{5+1/n}}.$$

Thank you all for the help. I have been working on this problem for more than 30 hours, I'm sure that together we can solve it.

  • $\begingroup$ You say it's not an exercise of this chapter, but is it an exercise or a question you ask yourself? $\endgroup$ – Davide Giraudo Nov 22 '11 at 23:50
  • $\begingroup$ @Kb100 : this is why I asked if you were sure of the space. $C^0([0,1])$ is not a hilbert space. $\endgroup$ – Glougloubarbaki Nov 23 '11 at 0:23

It is not closed.

Observe that $$ Kf(x) \;=\; \left(\int_0^1 f(y)\,dy\right)x^2 + 2\left(\int_0^1 y\,f(y)\,dy\right)x + \left(\int_0^1 y^2 f(y)\,dy\right) $$ In particular, the image of $K$ lies in the space of quadratic polynomials on $[0,1]$. The topology on this space given by the norm $\|\cdot\|_\infty$ for functions is the same as the standard topology on $\mathbb{R}^3$. Thus, it suffices to determine whether the set of triples $$ S \;=\; \left\{ \left(\int_0^1 f(y)\,dy,\;\int_0^1 y\,f(y)\,dy,\;\int_0^1 y^2 f(y)\,dy\right) : f\in \overline{B}([0,1])\right\} $$ is closed in $\mathbb{R}^3$.

To show that $S$ is not closed, consider functions $f$ for which $\|f\|_\infty \leq 1$ and $\int_0^1 f(y)\,dy=0$. If we do not require $f$ to be continuous, then the maximum value of $\int_0^1 y\,f(y)\,dy$ for such a function is $1/4$, which is attained for the function $$ f(x) \;=\; \begin{cases}-1 & \text{if } x<1/2, \\ 1 & \text{if } x \geq 1/2.\end{cases} $$ If we restrict to continuous functions then $1/4$ is not possible. However, it is possible to find a sequence of continuous functions $f_n\colon [0,1]\to[-1,1]$ so that $\int_0^1 f_n(y)\,dy = 0$ for all $y$ and $f_n\to f$ pointwise, in which case $\int_0^1 y\,f_n(y)\,dy \to 1/4$.


[Edit: This answer applies to an earlier version of the question, asking if an operator of the form $Kf(x) = \int_0^1 k(x,y) f(y)\,dy$ , where $k(x,y) = \sum_{i=1}^n a_i(x) b_i(y)$ with $a_i, b_i \in C([0,1])$, necessarily maps the unit ball of $C([0,1])$ to a closed subset of $C([0,1])$.]

I believe this is not true.

Take $k(x,y) = \cos(\pi y)$. Then $Kf$ is a constant function identically equal to $\int_0^1 f(y) \cos(\pi y)\,dy$. For $\|f\| \le 1$, this constant can be any number in $(-2/\pi, 2/\pi)$. (To get close to $2/\pi$, take a piecewise linear function equal to $1$ on $[0, \frac{1}{2}-\epsilon]$ and equal to $-1$ on $[\frac{1}{2}+\epsilon, 1]$). But the constant cannot equal $2/\pi$, since by continuity of $f$, either $f$ is strictly less than 1 on some interval in $[0,1/2]$, or strictly greater than -1 on some interval in $[1/2,1]$. So in fact the image of the unit ball under $K$ is not closed.

As mentioned by others, the Exercise 5.6 you cite does not apply here, since $C([0,1])$ is not a Hilbert space.

So if the special case addressed in the revised question ($k(x,y) = x^2 + 2xy+y^2$) is in fact true, it will need to use something special about this particular function $k$.


The kind of proof I have in mind uses weak convergence and is therefore beyond first year undergrad level (so, not very satisfying). Also since it turns out this is false with the $C^0(0,1)$ space I shall work in the space $L^2(0,1)$ with its usual norm.

Roughly speaking, you pick a sequence $K f_n$ in the image, where all the $f_n$ have norm less than one, assume that $K f_n$ converges quadratically to some $g$ (i.e. $\int (K f_n-g)^2$ tends to zero) and you'd like to find a $f$ of quadratic norm less than one such that $K f=g$. I don't think the difficulty here is related to the form of $k$(for $k$ degenerate), but rather to the wide possible choices of $(f_n)$ (so I'd say the problem is about as hard even is $k$ is a sum of polynomials).

Now the trick is that even though the unit ball in the space of $L^2$ functions is not compact, you can find a subsequence of any sequence $(f_n)$ that converges in some sense to some $f$ in the unit ball (that result is way too difficult for first year undergrad. Just admit it). In more theoretic words, since $L^2$ is a separable Hilbert, its unit ball is weakly sequentially compact. To make it more precise : there is a $f$ such that for all $h \in L^2$, $\int f_n h \rightarrow \int f h$. Add to that the hypothesis that $K$ is degenerate and you have the right $f$.

To go back to your example $f_n(x)=\sin(n x)$, the $f_n$ do not converge uniformly or in quadratic mean to 0, but they do converge weakly to 0 since it can be shown that for any $h \in L^2$, $\int h(x) \sin(n x) dx \rightarrow 0$.

I don't think you can do without weak convergence here.

  • $\begingroup$ I appreciate the warning but I am very determined! You say there is some subsequence of $(f_n)$ which converges in some sense to some $f$ in the unit ball. Consider $f_n(x) = \sin(nx)$ as this example arose in a test case. Clearly no subsequence of $(f_n)$ converges in the usual sense, so in this particular case, what is meant by "in some sense"? And where can I go to read more about this type of convergence? $\endgroup$ – nullUser Nov 22 '11 at 23:14
  • $\begingroup$ I see that $f_n(x)=\sin(nx)$ converges in some sense to 0, which in fact was the correct answer in that case. For any fixed $x$ when $n$ is large it gets arbitrarily close to a multiple of $2\pi$ so it must be going to $0$. Is this what you mean? $\endgroup$ – nullUser Nov 22 '11 at 23:17
  • $\begingroup$ well first I didn't say that any sequence actually converges, but only a subsequence. second, are you sure your statement is actually true ? (sorry, I wrote my answer a bit too vaguely and overlooked some difficulties). in these questions the choice of the space and of the norm is crucial. (are you sure you need to work with the uniform norm ?) if you really want to try this, look up "weak convergence", but I warned you ! ;) $\endgroup$ – Glougloubarbaki Nov 22 '11 at 23:41
  • $\begingroup$ Yes I'm sure $C([0,1])$ is the correct space using the uniform (max) norm. I am not sure about the result in general, but it is definitely true for polynomials, though I don't have the proof. $\endgroup$ – nullUser Nov 23 '11 at 0:02
  • $\begingroup$ but did you see the result in a book or do you just believe it is true ? $\endgroup$ – Glougloubarbaki Nov 23 '11 at 0:08

OK. So to answer the new question :

if $f \in C^0(0,1)$ has norm less than one then $K f$ is a polynomial of degree 2 with its coefficients lying in some bounded set. So if the image $I$ is closed then it is compact (and sequentially compact since metric).

A closer analysis shows that the leading coefficient $\int_0^1 f(y)dy$ can belong to any value in $[-1,1]$. The second, $2 \int_0^1 y f(y)dy$, will be maximal for $f=1$ and minimal for $f=-1$, as will the last, $\int_0^1 y^2 f(y)dy$. So, to sum it up, I is included in the set of polynomials of degree 2 with the coefficient in $x^2$ belonging to [-1,1], coefficient in $x$ belonging to [-1,1] and constant coefficient belonging to [-1/3,1/3].

EDIT : $I$ cannot be equal to all those polynomials since for example if $\int f =1$ and $f$ is in the unit ball then $f$ must be identically equal to 1, and we can't have 2 $\int y f =-1$. So more work is required.

  • $\begingroup$ I should have seen that from the start... this should in fact be easy to generalize to any degenerate kernel operator $\endgroup$ – Glougloubarbaki Nov 24 '11 at 22:39
  • $\begingroup$ The main problem then is how do I find the polynomials or piecewise linear maps? $\endgroup$ – nullUser Nov 24 '11 at 22:50
  • $\begingroup$ OK so I'm fairly sure this works : try piecewise linear maps with changing slopes at, say, 1/3 and 2/3. now give a name to the values at 0, 1/3, 2/3 and 1 which completely determine such a function and solve for these value the system with a, b, c. the system should even be under determined, but you'll find solutions $\endgroup$ – Glougloubarbaki Nov 24 '11 at 23:06
  • $\begingroup$ maybe only one change of slope would work, actually $\endgroup$ – Glougloubarbaki Nov 24 '11 at 23:13
  • $\begingroup$ It cannot be as simple as this. For example, you cannot achieve $Kf(x) = -x+1$. (To get the constant term to be 1, you must take $f=1$, but then the linear term is $x$, not $-x$.) $\endgroup$ – Nate Eldredge Nov 25 '11 at 0:02

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