If $X$ is the space of all bounded continuous Real valued functions defined on $\mathbb{R}$ . For every $f$ belonging to $X$ define $$ \|f\|_\infty=\{\sup|f(x)| ,x \text{ belongs to } \mathbb{R}\}. $$ I have proved that $X$ is a Banach space. The question is: $Y=\{f \text{ belongs to } X , f \text{ is differentiable on } \mathbb{R}\}$, is $Y$ a closed subspace of $X$?
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 29 '22 at 08:54
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You usually consider that it's false in a real analysis class, here it's just written differently – Jakobian Oct 29 '22 at 10:34
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The conclusion is false. Let $$f(x)=\begin{cases} (1-x^2)^2(1-|x|) &|x|\le 1\\ 0 & |x|>1 \end{cases}$$ Then $f$ is continuous, bounded and not differentiable at $x=0.$ Indeed $f'_+(0)= -1$ and $f'_-(0)=1.$ By the Weierstrass theorem there is a sequence of polynomials $p_n(x)$ such that $p_n(x)\rightrightarrows 1-|x|$ for $|x|\le 1.$ Let $$f_n(x)=\begin{cases} (1-x^2)^2 p_n(x) & |x|\le 1\\ 0 & |x|>1 \end{cases}$$ Then $f_n$ is differentiable for $x\neq \pm 1.$ On the other hand $$(f_n)'_+(1)=0=(f_n)'_-(1)=0,\qquad (f_n)'_+(-1)=0=(f_n)'_-(-1)=0$$ Hence $f_n$ is differentiable. Moreover $f_n\rightrightarrows f.$
Ryszard Szwarc
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Elaborating on the great answer of Ryszard Szwarc. The convergence of sequences with the norm you have defined is called "uniform convergence" and in Ryszard answer's is denoted by the double arrow.
Then, to show that your set $Y$ is not closed we can show that $X-Y$ is open. Ryszard has shown that there exists a function $f\in X-Y$ such that every open neighbourhood contains elements of $X$ ($f_n$ for some $n$). So $X-Y$ cannot be open, and $Y$ is not closed.
A. J. Pan-Collantes
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