$||f||=\max\limits_{0 \leq x \leq 1}|x^2f(x)|< \infty$ is not complete?

Question

$f$ is continuous function on $[0,1]$ and $||f||=\max\limits_{0 \leq x \leq 1}|x^2f(x)|< \infty$

Indeed, this defines a norm.

Now, I need to show that this norm is not complete.

Can anyone help?

Answer 1

A thorough description of what happens when you put that norm on $C[0,1]$ is in this previous question. If you want a specific example we can do like @Cameron and @qbert suggest and use a function that blows up at $0$. Define $f_n$ as $$ f_n (x) = \begin{cases} n & x \in [0,1/n]\\ 1/x & x \in [1/n,1] \end{cases} $$

To prove $\{f_n\}$ is Cauchy assume that $ m \lt n$ and examine $ |x^2(f_n(x) - f_m(x))|$ (which we'll call $d$) on three different sub-intervals.

On $[1/m,1]$, $f_n$ and $f_m$ are identical, so $d= 0$.

On $[1/n,1/m]$, $d = |x^2f_n(x) - x^2f_m(x)| = |x^2\cdot1/x - x^2m| \le x + (1/m^2)\cdot m \le 2/m.$

On $[0,1/n]$, $d \le (n-m)/n^2 \le (n+n)/n^2 = 2/n.$

So $\|f_n -f_m\| \le 2\max( 1/n, 1/m)$ and we've established the Cauchy condition.

To show that $\{f_n\}$ has no limit in $C[0,1]$, assume $f_n \rightarrow f$. For any $x_0 \ne 0$, $$ x_0^2|f(x_0) - f_n(x_0)| \le \|f-f_n\|.$$ Leave $x_0$ fixed and let $n \rightarrow \infty$ and we get $$ f(x_0) = \lim_{n\rightarrow \infty} f_n(x_0) = 1/x_0.$$ But there is no $f \in C[0,1]$ that agrees with $1/x$ on $(0,1]$.

Answer 2

(This is a reply to Jacky Chong's answer, $f_n(x) = (1-x)^{2n}$.)

While $\lim_{n \rightarrow \infty} f_n$ converges pointwise to a function that is not continuous, I don't think this is a valid counter-example. With the regular $\sup$ norm $$|f(x_0) - f_n(x_0)| \le \|f - f_n\|_0$$ (for any fixed $x_0 \in [0.1]$), so the point-wise limit is the only possible norm-limit, but that isn't true for the specified norm when $x_0 = 0$.

I haven't done the calculations, but this Desmos graph strongly suggests that, in the given norm, $f_n \rightarrow {\bf 0}$, the function that is identically zero on $[0,1]$.

Answer 3

Indeed, this space, with this norm is not complete, but this requires a careful proof.

Let $$ f_n(x)=\left\{\begin{array}{ccc} \frac{1}{x} & \text{if} & x\in \big[\frac{1}{n},1\big], \\ n & \text{if} & x\in \big[0,\frac{1}{n}\big]. \end{array} \right. $$ It can be shown that, for $m\ge n$ $$ \|\,f_m-f_n\|\le\frac{1}{4n}. $$ Hence $\{f_n\}$ is a Cauchy sequence.

The next thing to show is the following. If there was an $\,f\in[0,1]$, such that $\|\,f_n-f\|\to 0$, then, $\,f(x)=1/x$, for all $x\in(0,1]$. Which guarantees that such an $f$ DOES NOT exist in $C[0,1]$. To see this, for every $a\in (0,1)$ and $\delta>0$, set $$ g_{a,\delta}(x)=\left\{\begin{array}{ccc} 0 & \text{if} & x\in [0,a-\delta], \\ \frac{1}{\delta}(x-a+\delta) & \text{if} & x\in [a-\delta,a], \\ \frac{1}{\delta}(a+\delta-x) & \text{if} & x\in [a,a+\delta], \\ 0 & \text{if} & x\in [a+\delta,1]. \\ \end{array} \right. $$ Then, if $\,h\in C[0,1]$, it is readily seen that $$ \lim_{\delta\to 0}\int_0^1 h(x)g_{a,\delta}(x)\,dx= h(a). $$ Next, observe that, for all $a>1/n$, we have that $$ \lim_{\delta\to 0}\int_0^1 f_n(x)g_{a,\delta}(x)\,dx= f_n(a)=\frac{1}{a}. $$ Finally, $$ \int_0^1 f(x)g_{a,\delta}(x)\,dx=\int_0^1 \big(f(x)-f_n(x)\big)g_{a,\delta}(x)\,dx+\int_0^1 f_n(x)g_{a,\delta}(x)\,dx, $$ but $$ \Big|\int_0^1 \big(f(x)-f_n(x)\big)g_{a,\delta}(x)\,dx\,\Big|\le \int_{a-\delta}^{a+\delta} \big|\,f(x)-f_n(x)\big|\,g_{a,\delta}(x)\,dx \\ \le \int_{a-\delta}^{a+\delta} x^2\big|\,f(x)-f_n(x)\big|\,\frac{1}{x^2}\,g_{a,\delta}(x)\,dx\le \|\,f_n-f\|\int_{a-\delta}^{a+\delta}\frac{g_{a,\delta}(x)\,dx}{x^2}\le \frac{\|\,f_n-f\|}{(a-\delta)^2}, $$ since $\int_0^1 g_{a,\delta}(x)\,dx=1$.

and $\,\int_0^1 \big(f(x)-f_n(x)\big)g_{a,\delta}(x)\,dx=0$, whenever, $a-\delta\ge 1/n$. Hence, for any $a\in(0,1)$ and $n$ sufficiently large, $$ \Big|\,f(a)-\frac{1}{a}\,\Big|\le \Big|\,f(a)-\int_0^1 f\,g_{a,\delta}\,dx\,\Big|+\Big|\int_0^1\big(f_n-f\big)g_{a,\delta}\Big|+\Big|\,\int_0^1 f_n\,g_{a,\delta}\,dx-\frac{1}{a}\,\Big| $$ and since the RHS can become arbitrarily small, for suitably large $n$ and suitably small $\delta$, then $f(a)=1/a$.

Answer 4

Here is a slightly more general, albeit abstract approach.

Theorem. If $\varphi $ is a continuous function on $[0,1]$ such that the set $$ Z= \{x\in [0,1]: \varphi (x) = 0\} $$ has empty interior, define $$ \|f\|_\varphi = \sup_{0\leq x\leq 1}|\varphi (x)f(x)|, $$ for $f\in C([0,1])$. Then $\|\cdot\|_\varphi $ is a norm. It is complete iff $Z$ is empty.

Proof. Consider the linear transformation $$ T: f\in C([0, 1]) \mapsto \varphi f \in C([0, 1]). $$ Notice that $T$ is an injective map because if $T(f)=0$, then $f$ must vanish on $[0, 1]\setminus Z$, but since this set is dense, one must have that $f=0$.

Observing taht $$ \|f\|_\varphi = \|T(f)\|, \quad \forall f\in C([0, 1]), \tag 1 $$ it is clear that $\|\cdot \|_\varphi $ is a norm.

Regarding the last sentence we will just prove the "only if part" since the converse is much easier.

It follows from (1) that $\big (C([0, 1]), \|\cdot \|_\varphi \big )$ is isometrically isomorphic to the range of $T$. Assuming that $\|\cdot \|_\varphi $ is complete, we deduce that the range of $T$ is also complete, and hence closed in $C([0, 1])$.

By the Open Mapping Theorem, we then conclude that there exists a constant $K>0$ such that $$ \|T(f)\|\geq K\|f\|, \quad \forall f\in C([0, 1]). $$ Assuming by contradiction that $Z$ is nonempty, choose a point $x_0$ in $Z$, as well as a neighborhood $U$ of $x_0$, such that $|\varphi (x)|<K/2$, for $x$ in $U$. Choosing a continuous function $f$ supported on $U$, with $\|f\|=1$, we then have that $$ K/2\geq \|f\varphi \|= \|T(f)\|\geq K\|f\| = K, $$ a contradiction. QED

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Needless to say, if $\varphi (x)=x^2$, if follows from the above that $\|\cdot\|_\varphi $ is not complete.