Why do every strong extremum is simultaneously the weak extremum? ![enter image description here]( **My Doubt** > Here $||f||_{1}=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|$ where as $||f||_0=\sup_{x\in[0,1]}|f(...--prophetes.ai

Why do every strong extremum is simultaneously the weak extremum? ![enter image description here]( My Doubt > Here $||f||_{1}=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|$ where as $||f||_0=\sup_{x\in[0,1]}|f(x)|$. We can easily prove from definition that > > $$||f||_0=\sup_{x\in[0,1]}|f(x)|\leq \sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|=||f||_1$$ Using the above inequality, suppose $y_1$ is a strong extremum $\implies \exists \epsilon>0: ||y-y_1||_0<\epsilon$. How do we prove that it is a weak extremum? How is the underlined statement true?

If $\|f\|_1<\varepsilon$, then $\|f\|_0<\varepsilon$ must hold.