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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 $$\Delta$$.

cactus314
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