Dealing with exponential operators

Question

I have some troubles with the following equations. First, we define a function $g\left(\hat{\vec{x}}\right) = \exp[-(\vec{x}-\hat{\vec{x}})^2]$. We want to expand $g$ around $\hat{\vec{x}}\approx0$.

My first question is: What does it mean to expand an operator around zero? For a real number, I can imagine a small displacement perfectly well, but struggle to do so for an operator.

Next, the expansion is given to first order by $$g\left(\hat{\vec{x}}\right) \approx g(0) + \nabla g\left(\hat{\vec{x}}\right)|_{\hat{\vec{x}} = 0} *\hat{\vec{x}}$$

My next question is: How do I calculate this gradient? Of course, I first do exponential and square derivative, but how do I then evaluate $$\nabla(\vec{x}-\hat{\vec{x}})$$Shouldn't this then be a tensor? For example, we study stuff like this when looking at a Jacobian. However, in the paper that I am studying, the result is given as a dot product, so the result should indeed be a vector.

Answer 1

First of all I'm going to simplify the notation to save myself typing in latex. I replace the notation $\vec{x}$ for simply $x$ and $\hat{\vec{x}}$ for simply $y$.

Our functional (operator) $g: \mathbb{F}(\mathbb{R}) \mapsto \mathbb{F}(\mathbb{R})$ is defined by $$ g(y) = \exp\left[-\left(x - y\right)^2\right] $$

such that if we computed $g(0)$ we would get the function $f(x) = \exp\left[-x^2\right]$.

What does it mean to expand an operator around zero?

We want to approximate the functional $g(y)$ for $y \in \mathbb{F}(\mathbb{R})$ by finding some (hopefully simpler) functional $h: \mathbb{F}(\mathbb{R}) \mapsto \mathbb{F}(\mathbb{R})$ and computing $h(y)$.

Now $h$ will be a bad choice if the difference

$$ \lVert g(y) - h(y) \rVert $$

is not small.

How do i compute this gradient?

Let's assume:

• a) The gradient (now called a Ferchét derivative) exists
• b) That it is in fact $$ \nabla g(y) = h(y) = \exp\left[ -(x - y)^2 \right] \cdot 2 (x - y)$$

While we must produce this "qualified guess" of the estimate by sheer magic, we can check whether it is the Frechét derivative by asserting that $$ \lim_{\lVert \eta \rVert \rightarrow 0} \frac{\left \lVert g(y + \eta) - g(y) - h(\eta)\right \rVert}{\left \lVert \eta \right \rVert} = 0 $$

Here arises the first problem. While we know that $\mathbb{F}(\mathbb{R})$ is a vector space, we cannot compute the limit without knowing the norm $ \lVert \cdot \rVert $. This should be defined in the problem.

For some simplicity, we will choose the L2 norm,

$$ \lVert y \rVert_2^2 = \int_{-\infty}^{\infty} y(x)^2 dx $$

Now we can compute the limit $$ \lim_{\lVert \eta \rVert \rightarrow 0} \frac{\left \lVert g(y + \eta) - g(y) - h(\eta)\right \rVert}{\left \lVert \eta \right \rVert} $$ $$\lim_{\lVert \eta \rVert \rightarrow 0} \frac{\sqrt{\int_{\infty}^{\infty} \left(g(y + \eta) -g(y) - h(\eta) \right)^2 dx}}{\sqrt{\int_{\infty}^{\infty} \eta^2 dx}} $$

If we can show that the limit is zero, we have found our Frechét derivative.