# Seminorm exercise

Can you tell me if my answer is correct? It's another exercise suggested in my lecture notes.

Exercise: Consider $C[-1,1]$ with the sup norm $\|\cdot\|_\infty$. Let $$W = \{f \in C[-1,1] \mid \int_0^1 f d\mu = \int_{-1}^0 f d \mu = 0 \}$$

Show that $W$ is a closed subspace. Let $f(x) = x$ and calculate $\|f\|_{V/W} = \inf_{w \in W} \|f + w \|_\infty$ and show that the infimum is not achieved.

To show that $W$ is closed we show that if $f$ is a limit point of $W$ then $f \in W$. So let $f$ be a limit point. Then there is a sequence $w_n \in W$ converging to $f$, i.e. for $\varepsilon > 0$ there is $w_n$ such that $\|f - w_n\|_\infty < \varepsilon$. Hence for $\varepsilon > 0$, $\int_{0}^1 f d \mu = \int_0^1 (f + w_n - w_n ) d \mu \leq \int_0^1 |f-w_n| + \int_0^1 w_n = \int_0^1 |f-w_n| \leq \|f-w_n\|_\infty \leq \varepsilon$. Let $\varepsilon \to 0$. Same argument for $[-1,0]$.

Now we compute the norm: $$\|x\|_{V/W} = \inf_{w \in W} \|x + w\|_\infty = \inf_{w \in W} \sup_{x \in [-1,1]} |x + w(x)|$$

$\|x + w\|_\infty$ is smallest for $w(x) = -x$. But $-x \notin W$.

I'm not so sure about the second part. Is this what is meant by showing that the infimum is not achieved? "$\|x + w\|$ is smallest for $w(x) = -x$" seems a bit... wobbly. Thanks for your help.

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Make up your post it is full of mistakes and typos, which I think your see – Norbert Jul 9 '12 at 9:08
@Norbert The only mistake I can spot on this page is in your comment. I have taken a screen shot in case I need evidence in the future. – Rudy the Reindeer Jul 9 '12 at 9:18
When you are proving $f\in W$ you are talking about $w_n$. What is $w$? Why don't you write limits of integration? And why $\Vert f+w\Vert_\infty\leq\varepsilon$. You only assumed that $\Vert f-w_n\Vert_\infty<\varepsilon$. – Norbert Jul 9 '12 at 9:24
This is the only way not to make mistakes, when you have no intuition in new subject. Well my comments are boring but I think they are necessary when you study FA for the first time. – Norbert Jul 9 '12 at 9:32
Not at all. By the way you can prove that $W$ is closed by the following way. $W$ is an intersection of kernels of continuous functionals $I_1(f)=\int_0^1 f d\mu$, $I_2(f)=\int_{-1}^0 f d\mu$. It is known that kernels of continuous functionals are closed, then does $W$ as intersection of closed sets. – Norbert Jul 9 '12 at 9:39

Your idea for the first part is correct but the inequalities you write are odd. Try $$\left|\,\int_0^1f\,\right|=\left|\int_0^1(f-w_n)\right|\leqslant\int_0^1|f-w_n|\leqslant\|f-w_n\|_\infty\to0,$$ and similarly for the interval $[-1,0]$.

Regarding the second part, one would like to use the function $w_0:x\mapsto x-\frac12\mathrm{sgn}(x)$ to approximate $u:x\mapsto x$ but, despite the fact that $\int\limits_{-1}^0w_0=\int\limits_0^1w_0=0$, one cannot because $w_0$ is not continuous at zero. Hence $w_0$ is not in $W$ but the task is to show that $w_0$ indeed provides the infimum $\|u\|_{V/W}$.

Note that $u(x)-w_0(x)=-\frac12\mathrm{sgn}(x)$ for every $x$ hence $\|u-w_0\|_\infty=\frac12$. Call $W_0\supset W$ the set of integrable functions $w$ such that $\int\limits_{-1}^0w=\int\limits_0^1w=0$. For every $w$ in $W_0$, $\int\limits_0^1(u-w)=\frac12$ hence there exists some $x\geqslant0$ such that $u(x)-w(x)\geqslant\frac12$. This proves that $\|u-w\|_\infty\geqslant\frac12$ for every $w$ in $W_0$, and in particular for every $w$ in $W$, hence $\|u\|_{V/W}\geqslant\frac12$.

Furthermore, for any $w$ in $W$, the condition $\|u-w\|_\infty=\frac12$ implies that $u(x)-w(x)\leqslant\frac12$ for every $x$ in $[0,1]$. Since $\int\limits_0^1(u-w)=\frac12$ and $u-w$ is continuous, $u(x)-w(x)=\frac12$ for every $x$ in $[0,1]$. Likewise, $u(x)-w(x)=-\frac12$ for every $x$ in $[-1,0]$. These two conditions are incompatible at $x=0$ hence there is no function $w$ in $W$ such that $\|u-w\|_\infty=\frac12$.

Finally, one can modify $w_0$ to get some function $w_\varepsilon$ in $W$ such that $\|u-w_\varepsilon\|_\infty\leqslant\|u-v\|_\infty+\varepsilon$ hence $\|u\|_{V/W}=\frac12$. For example, one can consider the unique $w_\varepsilon$ in $W$ which is affine on $x\leqslant-\varepsilon$ and on $x\geqslant\varepsilon$ and such that $w_\varepsilon(x)=-x/(2\varepsilon)$ on $|x|\leqslant\varepsilon$.

Edit: For the last step, one could try to use the approximation of $w_0$ by its Fourier series, that is, to consider $w_n(x)=-\sum\limits_{k=1}^n\frac{\sin(2k\pi x)}{\pi k}$. Unfortunately, due to Gibbs phenomenon, this choice leads to $\|w_n-u\|_\infty$ converging to $\frac12+a$ where $a\approx0.089490$, instead of the desired limit $\frac12$.

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Thank you for this answer, did. I guess the main problem is to see that $\|u\|_{V/W} = \frac12$. How did you see that? I don't know how to see from $$\inf_{w\in W} \sup_{x \in [-1,1]} |x - w(x)|$$ that $w(x)$ should be $x - \frac{1}{2} \operatorname{sgn}{(x)}$. – Rudy the Reindeer Jul 9 '12 at 11:03
Here is how. Since $u$ is discontinuous at $0$, I first tried to determine the best possible $w$ on $(-1,0)$ with integral $0$ and the best possible $w$ on $(0,1)$ one with integral $0$. For example, if $w\geqslant u-c$ on $(0,1)$ with $c\lt\frac12$, there is no chance the integral of $w$ on $(0,1)$ can be $0$. Once I got those two functions $w$, I saw they were not making a continuous function on (-1,1)$... et voilà. – Did Jul 9 '12 at 14:35 Merci! I assume you mean$w_0$at the beginning of your second sentence. Before I got your comment I'd asked Jonas in chat how to see$\|u\|_{V/W}$and he said that it's "obvious" if you know the Dirichlet conditions. Which I did not know, of course. – Rudy the Reindeer Jul 10 '12 at 5:32 Here is an explicit sequence of functions that approaches the infimum: Let$\phi_{-}(f) = \int_{-1}^0 f$,$\phi_{+}(f) = \int_0^{1} f$. (Both$\phi_{-}$and$\phi_{+}$are continuous, hence$W = \phi_{-}^{-1}\{0\} \cap \phi_{+}^{-1} \{0\}$is closed.) Now consider the sequence of functions $$w_n(x) = \left\{ \begin{array}{ll} x-\frac{1}{2}-\alpha_n & \mbox{if } x \in [\frac{1}{n},1] \\ (\frac{1}{n}-\frac{1}{2}-\alpha_n)nx & \mbox{if } x \in [0,\frac{1}{n}] \\ -w_n(-x) & \mbox{if } x \in [-1,0) \end{array} \right.$$ A rather tedious calculation shows that to have$\phi_{+}(w_n)=0$(and hence$\phi_{-}(w_n)=0$, by oddness), I need to pick$\alpha_n = \frac{1}{4 n -2}$. Furthermore, it is clear that$\|f-w_n\|_{\infty} = \frac{1}{2}+\alpha_n$, from which it follows that$\inf_{w \in W} \|f-w\|_{\infty} = \frac{1}{2}\$.

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Nice, thank you copper.hat. I'll read it in more detail later, I'm working on something else right now. – Rudy the Reindeer Jul 19 '12 at 9:23