Bounding the L-Infty Norm of a Diffble Fn

[laonious]

Hello,

I would appreciate any assistance with the following question: Suppose f \in C^2[-1,1] is twice continuously differentiable. Prove that

|f'(0)|^2 \leq 4 ||f||_\infty (||f||_\infty + ||f''||_\infty), where ||f||_infty is the standard sup norm. At first I thought Taylor expansion seemed promising, but I am pretty stymied.

Thank you!

 

[mathman]

Fix your latex expression.

 

[laonious]

Hello,

I would appreciate any assistance with the following question: Suppose f \in C^2[-1,1] is twice continuously differentiable. Prove that

|f'(0)|^2 \leq 4 ||f||_\infty (||f||_\infty + ||f''||_\infty), where ||f||_{\infty} is the standard sup norm. At first I thought Taylor expansion seemed promising, but I am pretty stymied.

Thank you!

Sorry, here's a fixed-up rewrite.

 

[namphcar22]

First, as the inequality is invariant under the transformations f \mapsto cf, x \mapsto -x, we may assume WLOG that f(0), \ f'(0) \ge 0, and ||f||_\infty = 1. Thus, we must prove
|f'(0)|^2 \le 4(1 + ||f''||_\infty) subject to ||f||_\infty = 1. Now, for all x \in (0, 1) we have
1 \ge f(x) \ge f(0) + f'(0)x - \frac{x^2}{2} ||f''||_\infty, so f'(0) \le \frac{1 - f(0)}{x} + \frac{x}{2} ||f''||_\infty. The RHS is minimized for x^2 = x^2_0 = \frac{2(1 - f(0))}{||f''||_\infty} and decreasing for 0 < x < x_0. If x_0 \le 1 then |f'(0)|^2 \le \frac{1}{x_0^2} \left( 1 - f(0) + \frac{x^2_0}{2} ||f''||_\infty \right)^2 = 2(1 - f(0))||f''||_\infty \le 4||f''||_\infty.
If x_0 > 1 then |f'(0)|^2 \le \left(1 - f(0) + \frac{||f''||_\infty}{2} \right)^2 \le 4.

In either case |f'(0)|^2 \le 4 + 4||f''||_\infty.

 

[laonious]

namphcar,
Thanks for the great response, it was quite helpful. Would you mind being a little more explicit in your last step?

It's not clear to me that
\left(1-f(0)+\frac{||f''||_{\infty}}{2}\right)^2\leq 4,
since a function of the form e^{kx} would have an arbitrarily large second derivative.
Thanks!

 

[namphcar22]

Remember that x_0^2 = \frac{2(1 - f(0))}{||f''||_\infty}.