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Consider the regression model $\hat{y} = x^T\beta + v$, where $\hat{y}$ is the prediction, $x$ is a feature vector, $\beta$ is a coefficient vector, and $v$ is the offset term. If $x$ and $\tilde{x}$ are feature vectors with corresponding predictions $\hat{y}$ and $\tilde{y}$, show that $|\hat{y} − \tilde{y}| ≤ \lVertβ\rVert \lVert x − \tilde{x}\lVert$.
Endquote
I tried to tackle it with:
$$\lVert x^T \beta - {\tilde{x}}^T \beta \rVert \leq \lVert \beta \rVert \lVert x - {\tilde{x}} \rVert $$
and squaring both sides got to:
$$ (x^T\beta)^2 - 2x^T\beta{\tilde{x}}^T\beta + ({\tilde{x}}^T\beta)^2 \leq {\lVert \beta \rVert}^2 x^Tx - {\lVert \beta \rVert}^2 2x^T \tilde{x} + {\lVert \beta \rVert}^2 {\tilde{x}}^T\tilde{x} $$
From there, I used:
$$ a \cdot b = \lVert a \rVert \lVert b \rVert \cos\theta $$
which allowed me to get rid of $\lVert \beta \rVert$ and ultimately get to:
$$(\lVert x \rVert cos {\theta}_{x\beta} - \lVert \tilde{x} \rVert cos {\theta}_{\tilde{x}\beta})^2 \leq (x + \tilde{x})^2$$
where ${\theta}_{x\beta}$ denotes an angle between vectors $x$ and $\beta$.
I don't know where to take it from here. Does anyone have any suggestions how can I finish this? Or alternatively, is there a better way to tackle this problem in the first place?