Exercise · ZW-CONVERGENCE-WITH-LIPSCHITZ-GRADIENT-010 · ★★★ Advanced

Let \( f(x)=2x^2 \) on \( \mathbb{R} \), with \( x_0=1 \) and step \( t=1/L \). What is \( x_1 \)?

Choose one option.

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  • D. \( 0 \)

Explanation

Here \( \nabla f(x)=4x \), so \( L=4 \) and \( t=1/4 \). Then \( x_1=1-\frac{1}{4}\cdot 4=0 \). Using \( L=2 \) would give \( x_1=-1 \), a typical mistake.

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Review the concept: Convergence with Lipschitz gradient

Result guaranteeing that gradient descent with a suitable step size converges for differentiable functions whose gradient \( \nabla f \) is Lipschitz continuous with constant \( L \).

\[f(x_k)-f^* \le \frac{L\|x_0-x^*\|^2}{2k}\]

More practice problems: convergence with lipschitz gradient

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