Proving the Norm of a Hilbert Space: Tips and Tricks for Success

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To prove the norm of a Hilbert space, the discussion begins by addressing the case when x equals zero, establishing that the supremum is zero, which aligns with the norm of zero. For non-zero x, the relationship between the norm and the supremum is explored, demonstrating that the norm is less than or equal to the supremum of the inner product divided by the norm of y. The challenge lies in proving the reverse inequality, with participants suggesting the use of the Cauchy-Schwarz inequality as a potential solution. The conversation reflects a common struggle in mathematical proofs, particularly in higher-level concepts. Engaging with these foundational principles is crucial for mastering Hilbert spaces.
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Homework Statement



Let H be a Hilbert space. Prove \Vert x \Vert = \sup_{0\neq y\in H}\frac{\vert (x,y) \vert}{\Vert y \Vert}


The Attempt at a Solution


First suppose x = 0. Then we have \sup_{0\neq y\in H}\frac{\vert (x,y) \vert}{\Vert y \Vert} = \sup_{0\neq y\in H}\frac{\vert (0,y) \vert}{\Vert y \Vert} = \sup_{0\neq y\in H}\frac{\vert 0 \vert}{\Vert y \Vert} = 0 = \Vert 0 \Vert.

Now suppose x \neq 0. Then \Vert x \Vert = \sqrt{(x,x)} = \frac{\sqrt{(x,x)} \cdot \sqrt{(x,x)}}{\sqrt{(x,x)}} = \frac{\vert (x,x)\vert}{\Vert x \Vert} \leq \sup_{0\neq y\in H}\frac{\vert (x,y) \vert}{\Vert y \Vert}.

Now I just can't do the reverse inequality. Any help is much appreciated.
 
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jbunniii said:
Cauchy-Schwarz?

omg how do i call myself a math major.

thank you.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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