Proof that $\{f\in C([-1,1]): f(x)=0~\forall x\leq 0\}$ is a closed subspace of $C([-1,1])$

Question

Let $X=C([-1,1])$ and define $Y=\{f\in C([-1,1]): f(x)=0~\forall x\leq 0\}$. On $X$ we have the scalar product $\langle f_1,f_2\rangle=\int_{-1}^1 f_1(x)\overline{f_2(x)}dx$. I want to show that $Y$ is a closed subspace of $X$.

I wanted to start as follows: Let us take $f_n\in Y$ a convergent subsequence, say $f_n\rightarrow f$ in $X$. Since $X$ is complete we know that $f\in C([-1,1])$. So it remains to show that $f(x)=0$ for all $x\leq 0$.

I wanted to assume that there exists $x\leq 0$ such that $f(x)\neq 0$ but somehow I don't see how to get a contradiction with this. Did I start in the wrong way?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$One simple solution: as $X$ has the topology of a metric space, we need to check that $f_n\to f$ in $L^2$-norm, with $(f_n)\subseteq Y$, implies $f\in Y$.

How I'd do this question:

If $f_n\to f$, I know that $\forall\varepsilon>0,\,\exists N\in\Bbb N,\,n\ge N$ implies: $$\int_{-1}^0|f_n(x)-f(x)|^2\d x\le\int_{-1}^1|f_n(x)-f(x)|^2\d x\lt\varepsilon$$But $f_n\equiv0$ on $[-1,0]$, as $f_n\in Y$, so: $$\int_{-1}^0|f(x)|^2\d x\lt\varepsilon$$This last inequality is independent of $n$; it holds for any $\varepsilon>0$. That implies $\int_{-1}^0|f(x)|^2\d x=0$.

Note: we actually only needed: "$\forall\varepsilon>0,\,\exists N\in\Bbb N,\,\|f_N-f\|^2<\varepsilon$", i.e. some subsequence of $(f_n)$ converges to $f$

As $|f|^2$ is a continuous positive function, it is well known that its integral is zero iff. it is identically zero. So $f(x)=0$ for all $x\le0$. $\blacksquare$

As Ryszard comments, we know $f(x)=0$ for almost every $x\le0$. How can we see this implies $f(x)\equiv0,\,x\le0$ by continuity?

One easy way to do this is by integrating (which is essentially what I am doing in the above argument). Another 'direct' way to do this:

$f$ is uniformly continuous on $[-1,0]$. Thus for all $\varepsilon>0$ I know that there exists $\delta>0$, $|x-y|<2\delta$ implies $|f(x)-f(y)|<\varepsilon$ for $x,y\in[-1,0]$. Define the set $A\subseteq[-1,0]$ of points $x,\,f(x)\neq0$. We know it is Lebesgue-null, so there are intervals $(a_n,b_n)\subseteq[-1,0]$ with $A\subseteq\bigcup(a_n,b_n)$ and $\sum_n(b_n-a_n)<\delta$.

So let $x_0\in A$. I know $[-1,0]\cap(x_0-\delta,x_0+\delta)$ must contain some $y_0\notin A$: if not, then that'd imply $A\supseteq[-1,0]\cap(x_0-\delta,x_0+\delta)$, and then $\delta\ge\operatorname{diam}A\ge2\delta>0$, which is impossible. We have $|x_0-y_0|<2\delta$, so $|f(x_0)|=|f(x_0)-f(y_0)|<\varepsilon$ follows from the uniform continuity and from $f(y_0)=0$.

But then, for all $x_0\in A$, $|f(x_0)|<\varepsilon$ holds for every $\varepsilon>0$. This can only be true if $f(x_0)=0$ for all $x_0\in A$, but by definition of $A$ that is only true if $A=\emptyset$. So $f(x)=0$ for every $x\in[-1,0]$. $\blacksquare$

Can you generalise this to show: $$Y:=\{f\in X:f(x)=c,\,\forall x\in[a,b]\}$$Is closed in $X$ for any choice of $c\in\Bbb R,\,[a,b]\subseteq[-1,1]$?