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cactus314
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Show $L^2([0,1])$ with $|f|^2 = \int_0^1 (f(x)^2 + 0.5 \, f'(x)^2 ) \, dx$ is isomorphic to $L^2([0,1])$ with the standard norm

Let's consider two Hilbert spaces, copies of $L^2([0,1])$ with two different norms.

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

  • $H_2 = L^2([0,1])^2$ with norm $|f|^2 = \int_0^1 \big(f(x)^2 + 0.5 \, f'(x)^2\big) \, dx$

We need to show $H_1 \simeq H_2$ by finding a unitary transform.


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.


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.


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 $\Delta$.

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cactus314
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