Prob. 14, Sec. 1.5 in Erwine Kreyszig's Introductory Functional Analysis With Applications

Question

Let $C([0,1])$ be the set of all continuous real-valued function on $[0,1]$, let $d(f,g)=\int_0^1 |f(t) - g(t)|dx$.

We note that the exercise 14 needs exercise 13, which gives us a sequence $(f_n)_{n\in\mathbb{N}}$ of the form $$ f_n(t)= \left\{ \begin{array}{lcc} n & if & 0 \leq t\leq n^{-2}, \\ \\ t^{-1/2} & if & n^{-2} \leq t\leq 1. \end{array} \right.$$ which is a Cauchy sequence on $C([0,1])$. Show that this Cauchy sequence does not converge.

Here we are using the following ideas:

Suppose that this sequence is convergent, i.e, there exists $f\in C([0,1])$ such that $f_n$ converges to $f$. Then, we have $$\lim_{n\to+\infty}\int_0^\alpha |f_n(t)-f(t)|dt =0\quad\text{and}\quad \lim_{n\to+\infty}\int_\alpha^1 |f_n(t)-f(t)|dt =0$$ for all $0<\alpha<1$ and $\frac{1}{n^2}<\alpha$.

But let us observe that $$\lim_{n\to+\infty}\int_0^\alpha |f_n(t)-\infty|dt$$ does not exists and $$\lim_{n\to+\infty}\int_\alpha^1 |f_n(t)-t^{-1/2}|dt =0.$$

Therefore, there not exists $f\in C[0,1]$ such that the sequence $f_n$ converges.

Answer 1

(note that you are asking no question, so I'm just guessing what you want here)

As written, the argument makes no sense. You don't say what $\alpha$ is, and you seem to assume that the continuous function $f$ assumes the non-existent value $\infty$ on $[0,\alpha]$.

Instead, what you need to notice is that $f_n\to f$, where $f(t)=t^{-1/2}$, that limits in a metric space are unique, and that no continuous function on $[0,1]$ can agree with $f$ on $(0,1)$.

Answer 2

We have $f_k(t)=t^{-1/2}$ for $t\ge n^{-2}.$ If $f_n$ tended to $f$ we would have $f(t)=t^{-1/2}$ for $t\ge n^{-2}.$ As $n$ was arbitrary we would get $f(t)=t^{-1/2}$ for $t>0.$ The function $f$ is unbounded, hence it cannot be extended to a continuous function on $[0,1].$

Answer 3

Since $$\int_0^1\left|\frac1{\sqrt t}-f_n(t)\right|\,\mathrm dt=\int_0^{\frac1{n^2}}\left(\frac1{\sqrt t}-n\right)\,\mathrm dt=\frac1n,$$ if there was a function $f\in C[0,1]$ such that $d(f_n,f)\to0,$ we would have $$\int_0^1\left|\frac1{\sqrt t}-f(t)\right|\,\mathrm dt\le\frac1n+d(f_n,f)\to0$$ hence $$\forall\alpha\in(0,1]\quad\int_\alpha^1\left|\frac1{\sqrt t}-f(t)\right|\,\mathrm dt\le\int_0^1\left|\frac1{\sqrt t}-f(t)\right|\,\mathrm dt=0$$ i.e. $$\forall\alpha\in(0,1]\quad\forall t\in[\alpha,1]\quad f(t)=\frac1{\sqrt t},$$ i.e. $$\forall t\in(0,1]\quad f(t)=\frac1{\sqrt t}.$$ But there is no such function $f\in C[0,1].$