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Show (7.23)
We assume additive error model, so $Var(y) = \sigma^2_\varepsilon I$.
\[\begin{align*} \sum_{i=1}^N Cov(\hat{y_i}, y_i) &= tr[Cov(\hat{y}, y)] \\ &= tr[E[(\hat{y} - E[\hat{y}])(y - E[y])^T]] \\ &= tr[X(X^TX)^{-1}X^T \cancelto{Var(y) = \sigma^2_\varepsilon I_N}{E[(y - E[y])(y - E[y])^T]}] \\ &= d \sigma^2_\varepsilon \end{align*}\]where all expectations are over $y$.
See Weatherwax solution, contributed by Franklin Wang. I think the solution below uses same assumptions.
$\hat{f}(x_i)$ is $\hat{y_i}$. We can write the expected optimism as
\[\omega = E_\mathbf{y}[Err_{in} - \overline{\text{err}}] = \frac{1}{N} \sum^N_{i=1} E_\mathbf{y}[E_{Y^0}[(Y^0_i - \hat{y_i})^2] - (y_i - \hat{y_i})^2]\]using linearity of expectation to move the sum outside of $E_\mathbf{y}$. We need to show that the above is equivalent to
\[\frac{1}{N} \sum^N_{i=1} 2 Cov(\hat{y_i}, y_i) = \frac{1}{N} \sum^N_{i=1} 2 E_\mathbf{y}[(y_i - E_\mathbf{y}[y_i])(\hat{y_i} - E_\mathbf{y}[\hat{y_i}])]\]assuming that $y_i = f(x_i) + \varepsilon_i$, where $E[\varepsilon_i] = 0$ and $Var[\varepsilon_i] = \sigma^2$. Also, assume $Y_i^0 = f(x_i) + \varepsilon_i^0$. Errors are i.i.d.
In essence, there are two random variables, the training responses $\mathbf{y}$ and the new observation $Y_i^0$ at $x_i$ from $Err_{in}$. Randomness comes from the errors $\varepsilon$. All the variables except $f(x_i)$ are random as they are a function of $\mathbf{y}$ or $Y_i^0$.
Let’s show the equality of the i-th term:
\[E_\mathbf{y}[E_{Y^0}[(Y^0_i - \hat{y_i})^2] - (y_i - \hat{y_i})^2] = 2 E_\mathbf{y}[(y_i - E_\mathbf{y}[y_i])(\hat{y_i} - E_\mathbf{y}[\hat{y_i}])]\]For abbreviation, we will remove the i subscripts and write $f(x_i)$ as $f$.
First, note $E_{Y^0}[Y^0] = f + E[\varepsilon^0] = f$ and $E_{Y^0}[{Y^0}^2] = f^2 + 2 f E[\varepsilon^0] + E[{\varepsilon^0}^2] = f^2 + \sigma^2$. Then
\[E_{Y^0}[(Y^0 - \hat{y})^2] = E[{Y^0}^2] - 2 E[Y^0]\hat{y} + \hat{y}^2 = f^2 + \sigma^2 - 2f \hat{y} + \hat{y}^2\]So, LHS is
\[E_\mathbf{y}[(f^2 + \sigma^2 - 2f \hat{y} + \cancel{\hat{y}^2}) - (y^2 - 2y \hat{y} + \cancel{\hat{y}^2})] = \cancel{f^2 + \sigma^2 - E[y^2]} + 2 E[(y - f)\hat{y}]\]Adding $2 E[(y - f) E[\hat{y}]] = 2 E[(f + \varepsilon) - f] E[\hat{y}] = 0$, we obtain the RHS
\[2 E[(y - f)(\hat{y} - E[\hat{y}])] = 2 Cov(y, \hat{y})\]