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The Riesz's Lemma says, if $X$ is a Banach Space with norm $\|\cdot\|$ and $L$ is a closed subspace of $X$, then we have $$ \sup_{f:\|f\|=1} dist(f,L)=1, $$ where $dist(f,L)=\inf_{g \in L} \|f-g\|$.

As claimed by our professor, the following example demonstrates that the supremum might not be achieved in general: let $X=\{f \in C[0,1] \mid f(0)=0\}$ and $L =\{f \in X \mid \int_0^1 f = 0\}$.

I tried to prove this but have not found a clean way to do it, so I want to ask it here to see I could get some hints on this.

TTY
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So if you have $f\in X$ with $\|f\|=1$, you need to find some $h\in L$ with $\|f-g\|<1$.

There is some open subinterval $(a,b)\subset[0,1]$, in which $|f|<\frac12$.

Let $g\in X$ vanish outside $(a,b)$, with $g(x)>0$ for $x\in(a,b)$, and $\int_0^1g=\int_0^1f$, so that $f-g\in L$.

Now put $h=\varepsilon\cdot(f-g)$ with $\varepsilon>0$, and note that $\|f-h\|<1$ if $\varepsilon$ is small enough:

For $x\notin(a,b)$, $$|f(x)-h(x)|=|(1-\varepsilon)f(x)|<1-\varepsilon.$$ And for $x\in(a,b)$, $$|f(x)-h(x)|\le\tfrac12+\varepsilon(\tfrac12+\|g\|)<1 $$ if $\varepsilon$ is small enough.

Harald Hanche-Olsen
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  • Sorry to provide a full solution rather than just the hint you asked for. The solution seemed to write itself while I was figuring out how to nudge you in the right direction. – Harald Hanche-Olsen Oct 08 '14 at 18:20
  • @DavidMitra You do realize that the norm here is the maximum norm? I'm afraid I don't understand your comment. – Harald Hanche-Olsen Oct 08 '14 at 19:19
  • Sorry. I was completely off... – David Mitra Oct 08 '14 at 19:22
  • Thank you Harald! So basically you just want to use $\epsilon f$ as our $h\in L$ to get $|f-h|<1$, but $\int \epsilon f \neq 0$ so you have to subtract an $\epsilon g$ to get $h=\epsilon f - \epsilon g \in L$, and then to make sure $|f-h|=|(1-\epsilon)f+\epsilon g|<1$ by massage $g$ a little, very neat solution! – TTY Oct 08 '14 at 21:00
  • You got the basic idea, all right. Glad you liked it. – Harald Hanche-Olsen Oct 08 '14 at 22:37