No closest point to the subspace in $L_1([0,1])$

Question

Define $S_1=\{f\in L_1([0,1]),\int_{0}^1 xf(x) dx=0 \}$ . I want to show for every $\epsilon>0$, there exist $f$ in $S_1$ with $\lVert f-1\rVert_1 \leq1/2+\epsilon$, but there's no $f$ with $\lVert f-1\rVert_1 =1/2$. I got $\lVert f-1\rVert_1 \geq1/2$ from Holder's inequality, but not able to move further. Any help or hint is appreciated.

Answer 1

The inequality $\|f-1\|_1>1/2$ is immediate once you think about when the equality holds in Holder's inequality. That is $$\frac12=\left|\int_0^1x(f(x)-1)dx\right|\le\int_0^1x|f(x)-1|dx\le\int_0^1|f(x)-1|dx$$ with the equality if and only if $f(x)=1$ a.e.. If you are feeling unsure about this, we may consider the integral of non-negative function $(1-x)|f(x)-1|$, which is zero, implying $(1-x)|f(x)-1|=0$ a.e. and hence our claim.

But if $f(x)=1$ a.e., then $f\not\in S_1$ as $\int_0^1 f=1/2$. So there's no $f\in S_1$ such that $\|f-1\|=1/2$.

Example for the other direction can be constructed from our argument above. Define $$f_n(x)=\begin{cases} n^2/(n-1)^2,\quad &x\in[0,1-1/n]\\ -n/(2x),\quad&x\in(1-1/n,1] \end{cases}$$ It's straight-forward to verify $f_n\in S_1$. On the other hand, $$\|f_n-1\|_1=\left(1-\frac1n\right)\frac{2n-1}{(n-1)^2}+\frac1n+\frac n2\ln\frac n{n-1}$$ tends to $1/2$ as $n\to\infty$. In other words, for sufficiently large $n$ we have $\|f_n-1\|_1<1/2+\epsilon$, as is desired.