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I saw a proof of the non completeness of $N$ here. But intuitively, a proof for completeness could be:

Take $x_{n} \rightarrow x$ in $[a,b]$, and a Cauchy sequence $f_{n} \rightarrow f$ in $N$.

Using the continuity of $f_{n}$, and its convergence to $f$, we would get that, for every $\epsilon/3$, there is a $m$ such that for every $n > m$, we have $\left\lVert f(x_n)-f(x) \right\lVert < \left\lVert f(x_n)-f_n(x_n) \right\lVert + \left\lVert f_n(x_n)-f_n(x) \right\lVert + \left\lVert f_n(x)-f_n(x) \right\lVert < \epsilon$, so f is continuous on $N$, and thus it is a complete space.

Could you please help me understand what is wrong with this proof?

Thank you very much in advance.

Daniel Vasconcellos
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  • $f_n$ being Cauchy sequence if the norm $|\cdot|_1$ does not mean $f_n(x)$ is Cauchy for all $x.$ Also, I don't see how you can say $|f_n(x_n)-f_n(x)|<\epsilon/3$ for all $n>m.$ Look at the case $f_n(x)=x^n$ on $[0,1].$ Then consider the case when $x_n=1-1/n.$ Then $|f_n(x_n)-f_n(1)|\to1-e^{-1}.$ – Thomas Andrews Feb 11 '25 at 03:40

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You can't ensure $|f_n(x_n)-f_n(x)|<\frac \epsilon3.$ You've hand-waved why you can, so I can't say what is wrong with your reasoning for asserting that.

For example, if $[a,b]=[0,1]$ and $f_n(x)=x^n,$ then this sequence is Cauchy, and in fact converges to the zero function in $N.$ So it isn't even a counterexample.

But letting $x_n=1-1/n,$ then $f_n(x_n)=(1-1/n)^n\to e^{-1}$ and $f_n(1)=1,$ so $|f_n(x_n)-f_n(1)|\to 1-e^{-1}.$

Letting $x_n=1-1/\sqrt n$ would give $f_n(x_n)\to0.$

Your could make your proof work, I think, if you were using the $\|\cdot\|_\infty$ norm.

Thomas Andrews
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  • Your counterexamples are great. But shouldn’t continuity of $f_n$ imply that inequality? – Daniel Vasconcellos Feb 11 '25 at 04:36
  • No, you get $|f_n(x)-f_n(y)|<\epsilon/3$ iff $|x-y|<\delta_n,$ but there is no reason to,believe that the sequence converges fast enough for $|x_n-x|<\delta_n.$ Try to write out your argument in detail, and it will vanish in a puff of smoke. It is true that $|f_n(x_i)-f_n(x)|<\epsilon /3$ from all $i>m_n,$ for some $m_n.$ But there is no reason to expect $m_1,m_2,\cdots$ is a bounded sequence, and, in this example, we see than $m_n$ must be bigger than $n$ for all but finitely many $n$ when $\epsilon$ is small enough. – Thomas Andrews Feb 11 '25 at 04:47