Consider the subspace of continuous, real-valued functions on $[0,1]$ that are Lipschitz. Is this subspace complete under the sup norm ($\Vert \cdot \Vert_{\infty} = \sup \{ |f(x)| : x\in S \}$)?

I would say yes, since all Lipschitz functions ($d_{Y}(f(x_{1}),f(x_{2}))\leq K d_{X}(x_{1},x_{2})$, $K \geq 0$, where here $X = [0,1]$, $Y = \mathbb{R}$) are uniformly continuous, functions of certain types tend to converge uniformly to functions of the same types (e.g. differentiable functions to differentiable functions), but it seems unlikely.

Could somebody please help?

  • $\begingroup$ Is there inequality in the parentheses your definition of Lipschitz? If so then this would be a closed subspace of $C([0,1])$ and hence complete. Usually the definition of Lipschitz includes a constant and then the total space of Lipschitz functions would no longer be closed. $\endgroup$ – boxotimbits Dec 21 '15 at 4:57
  • $\begingroup$ @boxotimbits, yes, there should be a constant there. Wait while I fix it. $\endgroup$ – ALannister Dec 21 '15 at 4:58

No, the space is not complete, for the same reason that uniform convergence is not enough to preserve smoothness - after all, a uniform limit of smooth functions need not be even differentiable. For an example of just how horrible this can be, take the Weierstrass function; it's a uniform limit of $C^{\infty}$ functions that's nowhere differentiable.

For an explicit example, one can define a sequence of Lipschitz functions converging uniformly to $\sqrt{x}$; one construction is to take a piecewise linear function connecting $(0, 0)$, $(1/n, 0)$, $(2/n, \sqrt{2/n})$, and then $\sqrt{x}$ to the right of this. One easily checks that this converges uniformly to $\sqrt{x}$ (the pointwise difference is at most $\sqrt{2/n}$), but the Lipschitz norms blow up like $1/\sqrt n$.

If you don't like the corners, they can be smoothed out easily.

  • $\begingroup$ not taking Lipschitz norms, taking sup norms. $\endgroup$ – ALannister Dec 21 '15 at 5:00
  • $\begingroup$ @JessyCat Yes. The sequence is Cauchy using the sup norm, but not convergent - the limit function is not Lipschitz. The fact that the Lipschitz norms blow up doesn't prove this, but it does suggest how to prove the failure of convergence within your space. $\endgroup$ – user296602 Dec 21 '15 at 5:07

Theorem (Weierstrass). Any continuous $f:[0,1]\to R$ is the uniform limit of a sequence of real polynomials .....Now any continuously differentiable $g:[0,1]\to R$ is Lipschitz-continuous with Lipschitz constant $K=\max \{|g'(x)| :x\in [0,1]\}.$ Polynomials on $[0,1]$ are therefore Lipschitz. So with the $\sup$ norm, the set of Lipschitz-continuous $g:[0,1]$ is dense in $C[0,1]$, the space of all continuous $f:[0,1]\to R.$

  • $\begingroup$ Im not sure what you're getting at here. $\endgroup$ – ALannister Dec 21 '15 at 5:38
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    $\begingroup$ The space L of real Lipschitz functions on [0,1] with the sup norm is not complete because for any f in C[0,1] there is a Cauchy sequence of members of L, converging to f. And some members of C[0,1] are not in L. $\endgroup$ – DanielWainfleet Dec 21 '15 at 6:01

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