I'm trying to solve this question:

Let $$M=\{f\in C([0,1])\ | \ f(1)=0\}. $$

Is $M$ closed in $(C[0,1]),d_1)$?

At first, I assumed that $M$ was closed, so I tried to prove that if there is a sequence $f_1,f_2,f_3,...\in M,$and $\ f_n\rightarrow f\,$ with respect to $d_1$, then $f \in M$.

However, I got stuck and checked out the answers, and the answer I was given was just a counterexample of a sequence $(f_n)_n$ where $f_n \in M,\ f_n\rightarrow f$ wrt $d_1$, and $f\notin M$.

I understand that it's a lot easier to give a counterexample, but unfortunately I suck at analysis and cannot tell whether a set is closed or not upon inspection or come up with a counterexample right away. Are there other ways of proof, like proof by contradiction, to show that this set is not closed?

  • $\begingroup$ What is $d_1$? $d_1(f,g)=\int_0^1|f-g|$? $\endgroup$ – Lorenzo Quarisa May 5 '18 at 16:40
  • $\begingroup$ What metric are you using. Sup norm or some other norm $\endgroup$ – Foobaz John May 5 '18 at 16:40
  • $\begingroup$ @LorenzoQ. Yep that's the metric I'm using! $\endgroup$ – epsilonfox May 5 '18 at 16:48

A useful fact:

Let $X$ be a normed space and $\phi : X \to \mathbb{\mathbb{F}}$ a linear functional on $X$. Then $\ker \phi$ is a closed subspace of $X$ if and only if $\phi$ is bounded.


If $\phi$ is bounded, then $\ker \phi = \phi^{-1}(\{0\})$ which is closed as a preimage of a closed set under a continuous funciton.

Conversely, if $\phi$ is unbounded then there exists a sequence $(x_n)_n$ in $X$ such that $\|x_n\| = 1$ and $|\phi(x_n)| \ge n$ for all $n \in \mathbb{N}$.

Pick $x \in X \setminus \ker \phi$ and consider the sequence $\left(x - \frac{\phi(x)}{\phi(x_n)}x_n\right)_n$ in $\ker \phi$. Notice that it converges to $x$ which shows that $\ker \phi$ is dense in $X$. If $\ker \phi$ were closed, it would mean $\ker \phi = \overline{\ker\phi} = X$ so $\phi = 0$, contradicting the fact that $\phi$ is unbounded.

Consider the linear functional $\phi : X \to \mathbb{\mathbb{R}}$ given by $\phi(f) = f(1)$.

$\phi$ is not bounded. Namely, consider the sequence $(f_n)_n$ in $C[0,1]$ given by $f_n(x) = nx^{n-1}$. We have

$$\|f_n\|_1 = \int_0^1 nx^{n-1}\,dx = 1$$

but $\phi(f_n) = f_n(1) = n$.

Finally, notice that $M = \ker \phi$ and conclude that $M$ is not closed.

  • $\begingroup$ Apologies, but is there another explanation? I think this is beyond the scope of what we've learned so far, and I'm struggling to understand it. $\endgroup$ – epsilonfox May 5 '18 at 17:49
  • $\begingroup$ @statsfox Well, the easiest method would be to find a sequence $(f_n)_n$ in $M$ which converges to an element in $C[0,1]\setminus M$. An example is $$f_n(x) = \begin{cases} 1, & \text{if $x \in \left[0, 1-\frac1n \right]$} \\ n(1-x), & \text{if $x \in \left[1-\frac1n,1 \right]$} \end{cases}$$ which is the constant $1$ on $\left[0, 1-\frac1n \right]$ and then a line segment connecting $\left(1-\frac1n, 1\right)$ with $(1,0)$. It is easy to see that $f_n \xrightarrow{n\to\infty} 1$, which is not in $M$. $\endgroup$ – mechanodroid May 6 '18 at 14:49

If you want to proceed by contradiction, you can show that $M$ is dense in $(C([0,1]), d_1)$.

Given any $v \in C([0,1])$, let $f_n(x) = v(x)$ if $x < 1 - \frac{1}{n}$ and otherwise let $f_n$ be the line from $(1,0)$ to $(1-\frac{1}{n}, v(1-\frac{1}{n}))$. Then $d_1(f_n, v) \leq \frac{\sup{v}}{n}$. So $f_n \to v$.

So, if $M$ were closed, it would have to be all of $C([0,1])$, which is a contradiction.


I'll try to give an intuitive idea of why $M$ is not closed without providing a counterexample. If $M$ were closed, then for $f_n,f\in C([0,1])$ the assertion $$f_n\to f\text{ in }L^1(0,1) $$ would imply $$f_n(1)\to f(1) $$ But this is not true - convergence in $L^1$ only implies a.e. convergence of a subsequence (even if restricted to continuous functions - see here).


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