3
$\begingroup$

Let $(C[0,1],\lVert\cdot\rVert_{\infty})$ be the set of continuous functions in $[0,1]$, and consider $X=\{f\mid f(0)=0\}$ and $Y=\{f \in X\mid \int^1_0f(x)\,\textrm{d}x=0\}$, subspaces of $C[0,1]$.

Prove that for all $f \in X$ such that $\lVert f\rVert_{\infty}=1$ we have that $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \}<1$.

I proved that $Y$ is a proper closed subspace of $X$, thus is Banach because I also proved that $X$ is Banach.

We have that $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \} \leqslant \lVert f-0\rVert _{\infty}=1$.

Now if I assume that $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \} =1$, can someone help me to derive a contradiction?

$\endgroup$
  • $\begingroup$ I don't think arguing by contradiction will get you far. You should try to construct such a function (maybe first a discontinuous one). The basic idea should be that you want the function to be almost zero starting at some point, and different at the very beginning (where you have freedom coming from the fact that $f$ is nearly zero). $\endgroup$ – tomasz Feb 24 '17 at 14:10
3
$\begingroup$

Let $f(0)=0$ and $||f||_{\infty}=1$. Put $\beta=\int_0^1 f(x) dx$. Then $|\beta|<1$.

It is possible to find an explicit evaluation of $\inf\{||g-f||_{\infty} | g\in Y \}$. Note that for any $g\in Y$, $$ \left| \int_0^1 (f(x)-g(x)) dx \right| \leq ||g-f||_{\infty}. $$ This gives $$ \left|\int_0^1 f(x)dx \right| =|\beta|\leq \inf\{||g-f||_{\infty} | g\in Y \}. $$

To prove reverse inequality, we look for $g(x)$ in the form of $$g(x)=f(x)-h(x)\beta.$$ Then we require $$ h(0)=0, \ \ \int_0^1 h(x) dx =1, \textrm{ and } \ \sup_{x\in [0,1]}|h(x)\beta|<1. $$

In case $\beta=0$, we put $g(x)=f(x)$. Otherwise, we look for a continuous function $h(x)$ satisfying $$ h(0)=0, \ \ \int_0^1 h(x)dx=1 , \textrm{ and } \ \sup_{x\in [0,1]}|h(x)|<\frac1{|\beta|}. $$

Let $\epsilon>0$. By using $h(x)$ defined by $$ h(x) = x^{\epsilon} (1+\epsilon),$$ we obtain $$ h(0)=0, \ \ \int_0^1 h(x) dx = 1, \ \textrm{ and } \ \sup_{x\in [0,1]} |h(x)| = 1+\epsilon. $$ Then $||g-f||_{\infty} = \sup_{x\in [0,1]} |h(x) \beta| = (1+\epsilon)|\beta|.$

Since we may take $\epsilon$ arbitrarily small, we obtain the result $$ \inf\{||g-f||_{\infty} | g\in Y \} = \left|\int_0^1 f(x)dx \right|<1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.