Problem 6

Exercise 9.6 (Compute a few functional norms)   Let $X$ be the set of continuously differentiable functions $f : [-1,1] \to\mathbb{R}$ under the norm

$$\Vert f\Vert = \sup_{x\in[-1,1]} \vert f(x)\vert$$ (9.29)

Determine the boundedness and norms for the following functionals

$$\phi_1(f) = f(0) \quad \phi_2(f)=\int_{-1}^1\operatorname{sign}(x)f(x) \quad \phi_3(f)=f'(0) \quad \phi_4(f) = \sum_{n=1}^\infty \frac{f(1/n)}{2^n}$$ (9.30)

Proof. For $\phi_1$, the norm is bounded:

$$\Vert\phi_1\Vert = \sup_{\Vert f\Vert=1} \Vert\phi_1(f)\Vert \leq \sup_{\Vert f\Vert=1}\vert f(0)\vert \leq \Vert f\Vert=1$$ (9.31)

Equality is achieved for any $f$ satisfying $\vert f(0)\vert=\Vert f\Vert$.

For $\phi_2$, first simplify

$$\phi_2(f) = \int_{-1}^0 (-1)f(x) + \int_0^1 f(x) = \int_0^1 f(x)-f(-x)$$ (9.32)

$$\Vert\phi_2\Vert = \sup_{\Vert f\Vert=1} \Vert\phi_2(f)\Vert \leq \sup_{\Vert f\Vert=1}\int_0^1 \vert f(x)-f(-x)\vert\leq \Vert 2f\Vert=2$$ (9.33)

For equality, consider a Fourier series converging to the function

$$f(x) = \begin{cases} -1 \quad &x \in [-1,0]\\ 1 \quad &x\in [0,1] \end{cases}$$ (9.34)

Since the functional can be evaluated and equals 2 in the limit, this means the norm equals 2.

For $\phi_3$, we can see the functional is unbounded by considering the sequence of functions $f_n(x)=e^{-nx^2}$. In this case, $\Vert f_n\Vert=1$ and each $f_n$ is continuously differentiable. The derivative is $f_n'(x)=-ne^{-nx^2}$ and so

$$\sup_n \vert\phi_3(f_n)\vert = \sup_n \vert n\vert = \infty$$ (9.35)

For $\phi_4$,

$$\Vert\phi_4\Vert = \sup_{\Vert f\Vert=1} \Vert\phi_4(f)\Vert \leq \sup_{\Vert f\Vert=1}\sum_{n=1}^\infty \frac{\Vert f\Vert}{2^n}=1$$ (9.36)

If $f$ is constant, then

$$\Vert\phi_4(f)\Vert = \left\Vert\sum_{n=1}^\infty \frac{f}{2^n}\right\Vert=\Vert f\Vert$$ (9.37)

so the norm equals one. $\qedsymbol$