Show that there exists no $f_0\in X$ such that $||f_0||=1$ and $\text{dist}(f_0,Y)=1$

Question

Let $X=\{f\in C[0,1]:f(0)=0\}$ with the sup-norm and $Y=\{f\in X:\int _0^1 f=0\}$

Show that there exists no $f_0\in X$ such that $||f_0||=1$ and $\text{dist}(f_0,Y)=1$.

Suppose such a $f_0$ exists then $||f_0||=1\implies \sup \{f_0(x):x\in [0,1]\}=1$ and $\inf\{||f_0-f||:f\in Y\}=1$

Then $\forall f $ with $\int f=0$ we have $\sup\{ |f_0(x)-f(x)|,x\in [0,1]\}\ge 1$

Then $\forall f $ with $\int f=0$ there exists $x_0$ depending on $f$ such that $|f_0(x)-f(x)|\ge 1$

Also $\sup \{|f_0(x)|:x\in [0,1]\}=1\implies \exists x_0$ such that $f_0(x_0)=1$.

I am unable to use the above information to find such a $f_0$ does not exist.Can someone please help?

Answer 1

You can prove that if $\Vert f_0\Vert=1$, then there is $f\in Y$ with $\Vert f_0-f\Vert<1$. Below is a way to do it.

Since $f_0(0)=0$, then there is an interval $[0,\epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(\epsilon)|\leq1/2$.

In order to guarantee that the function $f$ we will construct will satisfy $\Vert f-f_0\Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $\delta$ (how small will be specified later). Let $n:[-1,1]\to[-\delta,\delta]$ be given by $$n(x)=\begin{cases} x+1-\delta&\text{ if }-1\leq x\leq -1+\delta\\ 0&\text{ if }-1+\delta\leq x\leq 1-\delta\\ x-1+\delta&\text{ if }1-\delta\leq x\leq 1\end{cases}$$ This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $x\in[\epsilon,1]$. You can then prove that $|f(x)-f_0(x)|\leq 1-\delta$ for all $x\in[\epsilon,1]$.

As long as $\delta<1/2$, we also have $f(\epsilon)=0$. I'll assume this from now on, because $\delta$ can be initially taken very small.

Now we need to define $f$ on $[0,\epsilon]$, in a way that $f\in Y$. We will define $f$ as a "spike". If we choose $c\in\mathbb{R}$, define $f:[0,\epsilon]\to\mathbb{R}$ in such a way that:

• $f$ is linear on $[0,\epsilon/2]$;
• $f$ is linear on $[\epsilon/2,\epsilon]$
• $f(\epsilon/2)=c$.

Then the integral $\int_0^\epsilon f(x)dx$ is simply the area of the triangle of basis $\epsilon$ and heigth $c$, so it is equal to $c\epsilon/2$.

We need to guarantee that $\int_0^1f(x)dx=0$. So we want that \begin{align*} \int_0^\epsilon f(x)dx+\int_\epsilon^1f(x)dx&=0\\ \epsilon c/2=-\int_\epsilon^1f(x)dx \end{align*} thus we choose $$c=-2\int_\epsilon^1f(x)dx/\epsilon$$

Note that

$$|c|\leq \frac{2}{\epsilon}\int_\epsilon^1|f(x)|dx\leq 2\delta/\epsilon$$ because of the definition of $f=n\circ f_0$ on $[\epsilon,1]$, and $|n|\leq\delta$. Just choose $\delta$ small enough (depending only on $\epsilon$) in such a way that $|c|<1/4$.

Repeating all we did so far: as long as we take $\delta$ small enough (namely, $\delta<1/2$ and $\delta<\epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that

• $f\in Y$ (this is precisely why we take $c$ as above).
• for all $x\in[\epsilon,1]$, $|f(x)-f_0(x)|\leq 1-\delta$ (this is a computation we did above);
• for all $x\in[0,\epsilon]$, $$|f(x)-f_0(x)|\leq |f(x)|+|f_0(x)|\leq 1/4+1/2=3/4$$ (this is the choice of $\epsilon$ and $c$.)

The second and third items above imply that $\Vert f-f_0\Vert\leq \max(1-\delta,3/4)$, which is smaller than $1$.

Answer 2

Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:X\to\mathbb{R}$ be the functional $I(f)=\int_0^1 f$. Note that $Y=\ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/Y\cong\mathbb{R}$), with the usual quotient norm $\|f+Y\|=\operatorname{dist}(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=c\operatorname{dist}(f,Y)$$ for all $f\in X$.

Now, if we can just prove that $c\geq 1$, we're done. Indeed, that implies that $\operatorname{dist}(f,Y)\leq |I(f)|$ for any $f\in X$. If $\|f\|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $f\in X$ with $\|f\|=1$ we conclude that $\operatorname{dist}(f,Y)<1$.

So we just need to prove that $c\geq 1$. But since the same constant $c$ works for all $f\in X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $\epsilon>0$, a function $g_\epsilon\in Y$ such that $\|f-g_\epsilon\|\leq 1/2+\epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)\neq 0$, and then modify this function slightly to get the desired $g_\epsilon$ (let $g_\epsilon(0)=0$ and then have $g_\epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_\epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_\epsilon$ balance out to $0$).

(In fact, $c=1$, so $\operatorname{dist}(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $c\leq 1$, which is easier.)