Riesz Lemma:
Let $Y$ be a closed proper vector subspace of a normed space $(X, \lVert. \rVert)$ and let $0 < \epsilon < 1$. Then there exists $u_{\epsilon} \in X$ with $\lVert u_{\epsilon} \rVert =1$ such that $\lVert u-y \rVert \geq \epsilon$ for all $y \in Y$
Now I saw a counterexample for $\epsilon=1$, which I don't really understand:
Define $X:=\{x \in C[0,1] : x(1)=0 \}$ with the norm $\lVert \cdot\rVert_{\infty} $ and consider $Y:=\{x\in X: \int_0^1 x(t)dt=0 \}$.
Let us assume that riesz lemma does hold for $\epsilon=1$ ,then there exists $x \in X$ such that $\lVert x-u\rVert_{\infty}$ for all $u \in Y$
Now defne $x_n(t):=1-t^n \in X$ and
$\lambda_n:=\frac{\int x(t)dt}{1-\frac{1}{n+1}}$
Consider the sequence $u_n:=x-\lambda_n x_n$
Because of the way we chose $\lambda_n$ we know that $u_n$ is in $Y$. Thus $|\lambda_n|\geq 1$, which leads to a contradiction because $|\int_0^1 x| <1$
My Questions:
Since I only know that there exists $x \in X$ such that $\lVert x-u \rVert_{\infty} \geq 1$ I don't see why $u_n$ is in $Y$.
If $u_n$ is in $Y$ then:
$\int_0^1 u_n=0$ thus
$\int_0^1( x(t)- \frac{\int x(t) dt}{1- \frac{1}{n+1}}(1-t^n) dt)=0$
But the only thing I know about $x(t)$ is that $x(1)=0$ (since $x(t) \in X$). So why do we know that $u_n$ is in $Y$?
At the end of the proof it was mentioned that $|\int_0^1 x|<1$, where does this come from?
Best Answer
$$\int_0^{1}u_n(t)dt=\int_0^1( x(t)- \frac{\int_0^{1} x(t) dt}{1- \frac{1}{n+1}}(1-t^n) dt)$$ $$=\int_0^{1}x(t)dt-\int_0^{1}x(t)dt \int_0^{1}(1-t^{n})dt/(1-\frac 1 {n+1}).$$ Just compute $\int_0^{1}(1-t^{n})dt$ to see that $\int_0^{1}u_n(t)dt=0$.
For the second question recall that $\|x\|=1$. So $|\int_0^{1}x(t)dt| \leq \int_0^{1} |x(t)|dt \leq \int_0^{1} 1dt=1$. If this were an equality then we would have $|x(t)|=1$ for all $t$ but $x(1)=0$. Hence, $|\int_0^{1} x(t)dt|<1$.