Image of closed ball under degenerate integral operator is a closed set

Question

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.

Answer 1

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 $\int_0^1 y\,f_n(y)\,dy \to 1/4$.

Answer 2

[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 $k$.

Answer 3

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.

Answer 4

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.