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.
My answer:
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.