# $\{ f\in C[0,1]:\exists K: |f(x)|\leq Kx\}$ is closed

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 $$A$$?

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

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 $$C([0, 1])$$.
• 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? Nov 4, 2023 at 13:28
• 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 $x \in [0, \varepsilon_k]$. Nov 4, 2023 at 13:35