# Show that $C^1[a,b]$ with the uniform norm is not a Banach space

Consider the normed vector space $$(C^1[a,b],||\cdot||_{\infty})$$, we have to prove that it is not Banach.

From theory a space is Banach if it is complete, ie if every Cauchy sequence in the space converges to an element of the same space.

So what we have to do is:

• find a Cauchy sequence in the space
• show that the squence does not converge to an element of the space

Consider $$[a,b]=[-1,1]$$ and the sequence defined by $$f_k(x)=\sqrt{x^2+1/k},\ k\in\mathbb{N}$$. Since the argument of the root is always non-negative, the sequence is always differentiable and both $$f_k$$ and $$f'_k$$ are continuous. So $$\{f_k\}\in C^1[-1,1]$$.

Intuitively I see that $$\lim_{k\rightarrow+\infty} f_k(x)=\sqrt{x^2}=|x|\notin C^1[-1,1]$$, but I first have to show that the sequence is Cauchy.

From theory a sequence $$\{x_n\}$$ is Cauchy iff $$\forall\epsilon>0\ \exists N\in\mathbb{N}:\forall m,n>N$$ we have $$||x_n-x_m||<\epsilon$$.

In our case we have $$||\sqrt{x^2+1/n}-\sqrt{x^2+1/m}||\le||\sqrt{x^2+1/n}||+||\sqrt{x^2+1/m}||$$ (*).

The sum (*) should be less than $$\epsilon$$, so each term should be less than $$\epsilon/2$$.

Moreover, since we are working with the uniform norm and $$||f||_{\infty}=max\{|f(x)|:x\in [a,b]\}$$, I think that $$||f_k(x)||_{\infty}=\sqrt{2}$$ because $$max\{x^2:x\in [-1,1]\}=1$$ and $$max\{1/k:k\in\mathbb{N}\}=1$$. So the sum (*) should be less or equal to $$2\sqrt{2}$$.

How can I use these information to prove that $$\{f_k\}$$ is a Cauchy sequence?

• You don't want to throw away all the control by using that version of the triangle inequality. Try graphing the functions and see for what value(s) of $x$ the functions $f_n$ and $f_m$ are farthest apart. – Ted Shifrin Oct 25 '18 at 22:16
• I think is $0$ the value of $x$ for which $f_n$ and $f_m$ are farthest apart. – sound wave Oct 25 '18 at 22:28

Applying the triangle inequality like this is too coarse. We can proceed like this:

$$\|\sqrt{x^2+{1/n}}-\sqrt{x^2+{1/m}}\|=\left\|\frac{1/n-1/m}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\|= \\ = |1/n-1/m|\cdot\left\|\frac{1}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\|$$

Now, $$|1/n-1/m|=\frac{|n-m|}{nm}$$, and

$$\left\|\frac{1}{\sqrt{x^2+{1/n}}+\sqrt{x^2+{1/m}}}\right\| = \frac{1}{1/\sqrt n+1/\sqrt m} = \frac{\sqrt{nm}}{\sqrt n + \sqrt m}$$

(the denominator is easily seen to be at minimum for $$x=0$$).

So, assuming $$N, the product is

$$\frac{|n-m|}{\sqrt{nm}(\sqrt n + \sqrt m)} \leq \frac{m}{\sqrt{nm}(\sqrt n + \sqrt m)}=\frac{1}{\sqrt{n}(\sqrt{n/m}+1)}\leq \\ \leq \frac{1}{\sqrt{N}}$$

(since $$\sqrt{n/m}+1 \geq 1$$).

Solving $$\frac{1}{\sqrt N}<\varepsilon$$ gives $$N > \frac{1}{\varepsilon^2}$$. Thus, for

$$n,m > \frac{1}{\varepsilon^2}$$

we get

$$\|\sqrt{x^2+{1/n}}-\sqrt{x^2+{1/m}}\| \leq \varepsilon$$

• @GiacomoTabarelli $1/\sqrt N$ is not a constant - it depends on $N$. – lisyarus Oct 26 '18 at 12:33
• @soundwave I updated the answer. – lisyarus Oct 26 '18 at 12:35
• @soundwave The definition of a Cauchy sequence is "$\forall \varepsilon \exists N$ such that...", so you have to provide such $N$, given arbitrary $\varepsilon$. You are right - in practice this means finding a bound for $\|f_n-f_m\|$ that goes to zero as $N$ goes to infinity. – lisyarus Oct 26 '18 at 13:15
• @soundwave The problem is that $N$ is a natural number, so we can't really say $N=\varepsilon^2$. I've slightly fixed the answer so that it is less confusing. – lisyarus Oct 26 '18 at 14:03
• @soundwave Well, we cannot say anything in this case. E.g. for $\varepsilon=\frac{1}{2}$ there is no $N$ such that $\|f_n-f_m\| < \varepsilon$ - we only now that $\|f_n-f_m\| < \sqrt N$, and $\sqrt N$ is at least $1$. – lisyarus Oct 26 '18 at 14:19