Is $L^2([0,1])$ with $|f|^2 = \int_0^1 (f(x)^2 + 0.5 \, f'(x)^2 ) \, dx$ isomorphic to $L^2([0,1])$ with the standard norm

Question

Let's consider $L^2([0,1])$. I want to put two different norms on this vetor space, say:

• $H_1$ is $L^2([0,1])$ with standard norm $|f|^2 = \int_0^1 f(x)^2 \, dx $

• $H_2$ is $L^2([0,1])$ with a norm like $|f|^2 = \int_0^1 \big(f(x)^2 + 0.5 \, f'(x)^2\big) \, dx$ that also controls the first dervivative

Ideally, I want to show that both are Hilbert spaces (of course for the former that's well-known) and also that $H_1 \simeq H_2$ by finding a unitary transform.

But I have difficulties to makes this work. Could anyone help me to show this, or if it does not work exactly as envisioned, what else along these lines would be true?

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Below some thoughts on the problem:

At least, in $L^2$ we have Fourier series. Let $\displaystyle f(\theta) = \sum_{n \in \mathbb{Z}} a_n e^{in \theta}$ then $\displaystyle f'(\theta) = \sum_{n \in \mathbb{Z}} n \, a_n e^{in \theta}$ (if we're allowed to differentiate term-wise. The norms should more or less look like this:

• $\displaystyle |f|_1^2 = \int_0^1 \big[ \sum_{n \in \mathbb{Z}} a_n e^{in \theta} \big]^2 \, dx = \sum_{n \in \mathbb{Z}} a_n^2 $
• $\displaystyle |f|_2^2 = \int_0^1 \big[ \big(\sum_{n \in \mathbb{Z}} a_n e^{in \theta}\big)^2 + 0.5 \big( \sum_{n \in \mathbb{Z}} n \, a_n e^{in \theta} \big)^2 \big] \, dx = \sum_{n \in \mathbb{Z}} (1 + 0.5 \, n^2) \, a_n^2 $

The Fourier series tells us that $f'(\theta)$ exists if $a_n = o(n^{-1})$. However, there are many ways to approximate functions in $L^2$ where derivatives shouldn't exist. We could have $f(\theta) = \theta$ and approximate with $f_N(\theta) = \frac{1}{N} \{ N \theta\}$ and let $N \to \infty$. Then $f_N(\theta) \to f$ in $L^2$ and yet $f'_N(\theta)$ is $0$ almost everywhere.

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If we want to take derivatives, perhaps we could try to use the Lebesgue differentiation theorem :

$$ f(x) = \lim_{|B| \to 0 } \frac{1}{|B|} \int f \, dx $$

However, this does not tell us how to (try to define) an $f'(x)$ in this setting. E.g. We'd like to say that $f'(\theta) = 1$ in the previous example.

One more possiblity is a Sobolev norm where we explicitly put in the premise that $f'(\theta)$ exists.

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Here is a function which is close to the line $f(x) \approx x$ in $L^2$ whose slope should be close to $1$. And is not differentiable. The slope is $f'(x) \approx 0$ or $1$.

Since $f(x)$ is not differentiable at these points we could try to approximate $\frac{d}{dx} \approx \frac{1}{\epsilon}[f(x+\epsilon) - f(x)]$ and call this operator $\Delta$.

Answer 1

This answer has been heavily edited based on the helpful comments. I apologize for not having thought enough before writing my previous answer.

As pointed out by others, the problem as stated does not make sense because you can’t take the derivative of an arbritratry $L^2$ function. We can modify the problem to make it solvable.

Problem: Find a unitary transform between $L^2[0,1]$ and $H_2=\{\sum_n a_n \exp(2\pi i n t)| \sum_n n^2 |a_n|^2<\infty, a_n\in \mathbb C\}$ equipped with inner product $$\langle f, g\rangle = \int_0^1 f(t)\bar{g}(t) + 0.5*f’(t)\bar{g’}(t)\;dt. $$

We can define a unitary transform $U$ as follows.

If $f\in H_2$ and $f(t) =\sum_n a_n e^{2\pi int}$, define $U:H_2\rightarrow H_1$ by $$(Uf)(t) = \sum_n a_n\sqrt{1+2\pi^2 n^2}\; e^{2\pi int}.$$

If we take $f(t) = \sum_n a_n e^{2 \pi i n t}$ and $g(t) = \sum_n b_n e^{2 \pi i n t}$ where $\sum |a^2_n| (1+ 2 \pi^2 n^2)$ and $\sum |b^2_n| (1+ 2 \pi^2 n^2)$ are both less than infinity, then

• $f$ and $g$ are in $H_2$,
• $||f||^2_{H_2} = \sum |a_n|^2 + 1/2* 4\pi^2 n^2 |a_n|^2 <\infty$,
• $||g||^2_{H_2} = \sum |b_n|^2 + 1/2* 4\pi^2 n^2 |b_n|^2 <\infty$,
• $\langle f, g \rangle_{H_2} = \sum a_n \bar{b_n} +\sum 2\pi^2 n^2 a_n\bar{b_n}, \mathrm{,\ and}$
• $ \langle Uf, Ug \rangle_{L_2}= \sum a_n \bar{b_n} (1+2\pi^2n^2). $

Many thanks to David Ullrich who pointed out a number of issues with the previous versions of this answer.

Answer 2

Let's define $$\|f\|_0=\left(\int_0^1 f(t)^2\,dt\right)^{1/2}$$ and $$\|f\|_1=(\|f\|_0^2+\|f'\|_0^2)^{1/2},$$leaving open for a second what sort of function $f$ must be, and for that matter ignoring for right now what we mean by that derivative.

Important Note Because of a certain technicality below I'm going to regard these as norms on spaces of functions with period $1$ instead of functions defined on $(0,1)$. (So for example $L^2$ below is a space of $1$-periodic functions.)

Unimportant Note I omitted the $0.5$ that appears in the definition in the question above. That clearly doesn't really matter, the $0.5$ just introduces needless complication.

Now $H_1=(L^2,\|\cdot\|_0)$ is certainly a Hilbert space. But in fact we cannot say $(L^2,\|\cdot\|_1)$ is a Hilbert space, since $\|f\|_1$ is obviously undefined for the typical $f\in L^2$.

In fact $\|\cdot\|_1$ is the standard norm on the Sobolev space often denoted $W^1$, $H_2=(W^1,\|\cdot\|_1)$ is a Hilbert space, and once we get a few things sorted out it will be clear how to define an isometry between these two Hilbert spaces. This raises the question of exactly what $W^1$ is; the definition requires a bit of care.

Speaking very informally, people say that $W^1$ is the space of $f$ such that $f,f'\in L^2$. That's not literally correct; taken literally it implies that $f$ is differentiable, and the space of differentiable functions is not complete under our norm.

In a not quite successful attempt to rehabilitate the question, the OP says "One more possiblity is a Sobolev norm where we explicitly put in the premise that $f'(\theta)$ exists." No, there's no explicit mention of $f'(\theta)$ in the definition of $W^1$, at least if $f'(\theta)$ is the pointwise derivative.

In fact $W^1$ is the space of $f$ such that $f,Df\in L^2$, where $Df$ is the derivative of $f$, in an appropriate sense. It's interesting that there are many apparently different ways to define what sort of $Df$ we need here.

People often say $f$ "has a weak derivative in $L^2$" if there exists $g\in L^2$ such that $$\int_0^1f\phi'=-\int g\phi$$for every infinitely differentiable $\phi$ with period $1$. Note that this is the same as saying that the derivative "in the sense of distributions" is an $L^2$ function.

Or: if I say "$f$ is differentiable in $L^2$" you might reasonably assume I meant that $f'$ exists and $f'\in L^2$. No, when I say that I mean that the differentiation happens in $L^2$:

For $h\ne0$ let's define $$\Delta_hf(t)=\frac{f(t+h)-f(t)}h$$ (this is where I want to talk about periodic functions instead of functions defined on $(0,1)$). So the ("ordinary" pointwise) derivative is just the function $f'=\lim_{h\to0}\Delta_hf$. We could consider the limit in the $L^2$ norm instead:

Def the function $f\in L^2$ is differentiable in $L^2$ if there exists $g\in L^2$ such that $$\lim_{h\to0}\int_0^1|g-\Delta_h f|^2=0.$$

Finally, recall that if $f$ is absolutely continuous then it is differentiable almost everywhere; in this case we will write $f'$ for the almost-everywhere-defined pointwise derivative.

And now we can state something interesting:

Something Interesting. Suppose $f\in L^2$. The following are equivalent:

(i) $f$ has a weak derivative in $L^2$.

(ii) $f$ is differentiable in $L^2$.

(ii') If we say $\tau_hf(t)=f(t+h)$ and define $F:\Bbb R\to L^2$ by $F(h)=\tau_hf$ then $F$ is differentiable (in norm).

(iii) $\sup_{h\ne0}\int_0^1|\Delta_h f|^2<\infty$.

(iv)$\sum n^2|\hat f(n)|^2<\infty$.

(iv') There exists $g\in L^2$ with $\hat g(n)=2\pi in\hat f(n)$.

(v) $f$ is absolutely continuous and $f'\in L^2$.

Also, if any of those conditions hold then the Fourier series for the weak derivative of $f$ can be obtained in the obvious way from the Fourier series for $f$.

Edit: Of course this cannot be literally true, since modifying $f$ on a set off measure zero does not affect the firstt four conditions but has a major effect on (v). Of course, as usual when, say, we say a function in $L^1$ is continuous, what (v) really means is

(v') $f=\tilde f$ almost everywhere, where $\tilde f$ is absolutely continuous and $(\tilde f)'\in L^2$.

I don't have a reference handy; this is stuff I worked out years ago. A lot of it is trivial:

Some of the proof. If you've read this far it's obvious that (iv) is equivalent to (iv'). (i) is equivalent to (iv) by standard stuff about distributions. (ii) implies (iii) is trivial, and (iv) implies (ii) is easy (write down what the Fourier series for $f'$ "should" be, and use (iii) plus DCT to show that that series actually gives a function witnessing (ii). (ii) implies (iv) is easy (write down the Fourier series for $\Delta_hf$.))

I don't see how to give a one-line hint for (v), but if I haven't connected them all yet you should be able to use the hints above to work out for yourself why (i) through (iv) are equivalent. That's as close to qed as II have time for right now. (no wait: Come to think of it (v) implies (i) is a standard fact about AC; follows from Fubini and the fact that $f(x)=f(0)+\int_0^x f'$.

Edit: Leaving out (v) is unfortunate, since it's the one you could imagine actually checking for a given $f$. Actually (v) is the one part that didn't seem immediately obvious, had to think about how to prove it:

Proof that the other conditions imply (v'). Assume the other conditions. Say $$f(t)=\sum a_ne^{2\pi i nt},$$so that $$g(t)=2\pi i\sum n a_ne^{2\pi int}$$is the weak derivative. Note that (iv) shows that the Fourier series for $f$ converges uniformly, hence after modifying $f$ on a set of measure zero the series converges uniformly to $f$. If you regard $\int_0^x g$ as the inner product of $g$ and $\chi_{[0,x]}$ and work out the Fourier coefficients of the characteristic function then Parseval shows that $$\int_0^x g=\sum a_n(e^{2\pi inx}-1)=f(x)-f(0).$$This shows that $f$ is absolutely continuous (and hence $\int_0^xf'=f(x)-f(0)=\int_0^x g$, so $f'=g$ almost everywhere).

Condition (iii) doesn't say anything explicitly about the derivative of $f$; it's included because it's interesting, showing that "differentiable in $L^2$" is equivalent to "Lipschitz in $L^2$" (the latter being defined by (iii)).

If we regard (iv') as part of (iv) then each condition except (iii) gives an interpretation of what one might mean by "$f'\in L^2$". With one of those interpretations, $W^1$ is the space of $f$ such that $f,f'\in L^2$.

And now $H_2=(W^1,|\cdot|_1)$ is a Hilbert space and

Easy Exercise There exist $c_n$ such that $\sum a_ne^{2\pi int}\mapsto\sum c_na_ne^{2\pi int}$ defines an isometry from $H_1$ onto $H_2$.

In case this is not yet long enough:

Note. Condition (v) implies in particular that $f(y)-f(x)=\int_x^y f'$, with $f'\in L^2$. Applying Cauchy-Schwarz gives a special case of the Sobolev Embedding Theorem:

Cor. $W^1\subset\text{Lip}_{1/2}$.

Of course as noted above where we pointed out that what (v) "really means" is (v'), a more careful statement of the corollary would be

Cor'. If $f\in W^1$ then $f=\tilde f$ almost everywhere with $\tilde f\in\text{Lip}_{1/2}$.

In fact people never worry about this, they state things like the Sobolev Embedding Theorem as in the first version of the corollary and trust the reader to understand the proper "interpretation"...