Problem: $\lambda: C[0,1] \to C[0,1]$ given by $\lambda (f) = (x\mapsto xf(x))$. Find the closure of $A=\lambda(C[0,1])$ wrt. $||\cdot ||_\infty.$

I could prove that $A \subseteq G=\{ f\in C[0,1]: \exists K: |f(x)|\leq Kx\} \subseteq \overline{A}$.

If I show that $G$ is closed then I'm done. I'm not sure if this holds at all.

Please lead me to the answer. Is $G=\overline{G}$ true? If not, what other $G$ should I choose to find the closure of $A$?

  • 3
    $\begingroup$ Let $f_n(x)=\min\{\sqrt x, nx\}$, Then each $f_n$ is $\in A$, but their uniform limit is not in $G$ $\endgroup$ Nov 4, 2023 at 12:41
  • $\begingroup$ This means $G$ is not closed, right? $\endgroup$ Nov 4, 2023 at 12:45

1 Answer 1


Here is the full solution. Read only the hint if you want to solve it on your own.

Hint: What condition should $g \in C([0, 1])$ satisfy in order to be in $\lambda(C([0, 1])$?

To find the answer, suppose first that $g = \lambda(f)$, i.e. $g(x) = x \cdot f(x)$ for all $x \in [0, 1]$. In this case $x \mapsto g(x)/x$ is a continuous function on $[0, 1]$. Conversely, if $x \mapsto g(x)/x$ defines a continuous function on $[0, 1]$, then $g$ is in the image of $\lambda$. Convince yourself that this is true if and only if $$ \lim_{x \to 0} \frac{g(x)}{x} $$ exists.

Now that we know the image of $\lambda$, let's find its closure. Another property all $g$ in $\lambda(C[0, 1])$ share in common is that they vanish at $0$, i.e. $g(0) = 0$. If $h \in C([0, 1])$ satisfies $h(0) = 0$, then this is not enough to guarantee the existence of the limit above, e.g. $h(x) = \sqrt{x}$. However, any such function can be approximated uniformly by a sequence $(g_k)_{k \in \mathbb{N}} \subset \lambda(C[0, 1])$. Simply define $g_k$ by $g_k(x) = \frac{h(\varepsilon_k)}{\varepsilon_k} \cdot x$ for $x \in [0, \varepsilon_k]$ and $g_k(x) = h(x)$ in the rest of the interval. Here $(\varepsilon_k)_{k \in \mathbb{N}}$ is a sequence tending to $0$. We are done, since $\{h \in C([0, 1]) : h(0) = 0 \}$ is closed in $C([0, 1])$.

  • $\begingroup$ I have the same exact construction. The part that is bothering me, is the uniform convergence of $g_k$. For that we need to show something like: $\forall \epsilon >0 \exists \delta : |\frac{h(\epsilon_k)x}{\epsilon_k}-h(x)|< \epsilon$ on $[0,\delta] \forall k\geq N$. This seems non trivial to me. How would you show that it is unfiromly convergent? $\endgroup$ Nov 4, 2023 at 13:28
  • 1
    $\begingroup$ Let $\varepsilon > 0$. Since $h(0) = 0$, there exists $\delta > 0$ such that $|h(x)| < \varepsilon$ for $x \in [0, \delta]$. Choose $k$ large so that $\varepsilon_k < \delta$. Then $$ |g_k(x) - h(x)| \le \frac{|h(\varepsilon_k)|}{\varepsilon_k}x + |h(x)| \le \frac{\varepsilon}{\varepsilon_k} \varepsilon_k + \varepsilon = 2\varepsilon $$ for all $x \in [0, \varepsilon_k]$. $\endgroup$ Nov 4, 2023 at 13:35
  • $\begingroup$ Thank you! This filled in the gap for me. Great work $\endgroup$ Nov 4, 2023 at 13:37
  • $\begingroup$ You're welcome! :) $\endgroup$ Nov 4, 2023 at 13:38

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