Equivalence of norms and the inner product in $C([-1,1])$

Question

In $\mathcal{C}([-1,1])$ we define:

$$\varphi_a(f,g)=\int_{-1}^{1} f(t)\overline{g(t)}dt+(2a^2-1)f(a)\overline{g(a)}, \ \ \ f,g\in\mathcal{C}([-1,1]), a\in[-1,1].$$

Let $\Lambda=\left\{a\in [-1,1]: \ \varphi_a \mathrm{ \ is \ an \ inner \ product}\right\}$. Compute the set $\Lambda$ and all pairs $(a,b)\in\Lambda\times\Lambda$, such that $\varphi_a$ and $\varphi_b$ set equivalent norms.

I know conditions to be the inner product, but I can't see how the parameter $a$ could ruin it. I think that the linearity should work, conjugate symetry too, only the 'positive-definitness' maybe, but it depends on so many things: on value of the integral, values $f(a)$,$g(a)$ and finally on the $sign\left\{ (2a^2-1)\right\}$, I don't know how to compute it.

Any hint how to start?

Answer 1

Let $a\in [-1,1]$. For the positive-definiteness, you need $\varphi_a(f,f)\geq 0$ for all $f\in \mathcal{C}([-1,1])$.

The hint is to look at these two cases:

• $a$ such that $2a^2 - 1 \geq 0$.

• $a$ such that $2a^2 - 1 < 0$. In this case, consider the function $f_a$ s.t:

  $f_a(a)=1$, $f_a(x)$ is zero when $|x-a| \geq \frac{(1 - 2a^2)}{10}$, is a straight line in the interval $\left[a-\frac{(1 - 2a^2)}{10}, a\right]$ going from $0$ to $1$, and a straight line in the interval $\left[a, a+\frac{(1 - 2a^2)}{10}\right]$ going from $1$ to $0$.

What can you say about $\varphi_a(f_a,f_a)$ in this case?

Now, for the equivalence of the norms. Assume $a\ne b$. We have to look at these two cases:

• $a, b$ such that $2a^2 = 1 = 2b^2$. As you said, it is obvious the norms are equivalent in this case.
• $a, b$ such that either $2a^2 - 1 > 0$ or $2b^2 - 1 > 0$.

In this case, asume, w.l.o.g., $2a^2 - 1 > 0$. We will prove the norms are not equivalent in this case. One way to do this is to consider the sequence of functions $f_{a,n}$ s.t:

$f_{a,n}(a)=n$, $f_{a,n}(x)$ is zero outside of the interval $\left[a - \frac{1}{n^2}, a +\frac{1}{n^2}\right]$, is a straight line in the interval $\left[a-\frac{1}{n^2}, a\right]$ going from 0 to $n$, and a straight line in the interval $\left[a, a+\frac{1}{n^2}\right]$ going from $n$ to $0$.

Notice the function $f_{a,n}^2$ is nonzero only in a region included in a rectange of base $\frac{2}{n^2}$ and height $n^2$, whose area is $2$. Hence, $\varphi_b(f_{a,n}, f_{a,n})=\int_{-1}^1 f_{a,n}^2(x)dx\le 2$ for large enough $n$ s.t. $b$ is outside of the interval $\left[a - \frac{1}{n^2}, a +\frac{1}{n^2}\right]$.

On the other hand, $\varphi_a(f_{a,n}, f_{a,n})=\int_{-1}^1 f_{a,n}^2(x)dx + (2a^2 - 1) n^2 > (2a^2 - 1) n^2$ which goes to $+\infty$ as $n\to +\infty$.

Hence, for any constant $C>0$, we can find some $n$ s.t. $\varphi_a(f_{a,n}, f_{a,n}) > 2C \ge C\varphi_b(f_{a,n}, f_{a,n}) $. Hence, the norms are not equivalent.