Vtyjkyuk

Define $S_1=fin 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 $lVert f-1rVert_1 leq1/2+epsilon$, but there's no $f$ with $lVert f-1rVert_1 =1/2$. I got $lVert f-1rVert_1 geq1/2$ from Holder's inequality, but not able to move further. Any help or hint is appreciated.

The inequality $|f-1|_1>1/2$ is immediate once you think about when the equality holds in Holder's inequality. That is
$$frac12=left|int_0^1x(f(x)-1)dxright|leint_0^1x|f(x)-1|dxleint_0^1|f(x)-1|dx$$
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 $fnotin S_1$ as $int_0^1 f=1/2$. So there's no $fin S_1$ such that $|f-1|=1/2$.

Example for the other direction can be constructed from our argument above. Define
$$f_n(x)=begincases
n^2/(n-1)^2,quad &xin[0,1-1/n]\
-n/(2x),quad&xin(1-1/n,1]
endcases$$
It's straight-forward to verify $f_nin S_1$. On the other hand,
$$|f_n-1|_1=left(1-frac1nright)frac2n-1(n-1)^2+frac1n+frac n2lnfrac nn-1$$
tends to $1/2$ as $ntoinfty$. In other words, for sufficiently large $n$ we have $|f_n-1|_1<1/2+epsilon$, as is desired.