I have a norm on space $X = C[0,2]$: $$p(x) = \max_{t \in [0,2]}|x(t)| + \left(\int_0^1|x(t)|^7\right)^{1/7}$$ Is that norm induced by any scalar product?
I try to find counterexample for parallelogram identity $$||f+g||^2 + ||f-g||^2 = 2(||f||^2 + ||g||^2)$$ Why I tried to find it? Because I know the norm $$d(x) = \max_{t \in [0,2]}|x(t)|$$ can't be induced by any scalar product. Maybe I should use polar identity: $$(f,g) = \frac{1}{4}(||f+g||^2 + ||f-g||^2)$$
And then what to do with the norm, where instead of the integral - the sum? For example on space $C[-1,1]$: $$k(x) = \max_{t \in [-1,1]}\pi^{2t}|x(t)| + 5\left(\sum_{k=1}^{\infty}\frac{\left|x(-1+\frac{3}{2k})\right|^2}{k^5}\right)^{1/2}$$
Here, too, nothing happened. I know that a sum can be obtained from the scalar product (if $p=2$), if we consider it as a separate norm. And the maximum is impossible. How to work with such norms, where one of two norm can be induced?
Can you tell me a trick or something? Thank!
P.S (Edit): Maybe there is a chance to somehow prove that it is impossible to induce any norm that already includes a norm that is not inducible by anything?
Consider
$$f(x) = \begin{cases} \cos \pi x & 0 \le x \le 1/2\\ 0 & 1/2 \le x \le 2 \end{cases}$$
and $g(x) = f(2-x)$. You have $f,g \in \mathcal C([0,2],\mathbb R)$. Let $$I = \left(\int_0^2|f(t)|^7\right)^{\frac{1}{7}}=\left(\int_0^2|g(t)|^7\right)^{\frac{1}{7}} = \frac{1}{2}\left(\int_0^2|f(t)+g(t)|^7\right)^{\frac{1}{7}}=\frac{1}{2}\left(\int_0^2|f(t)-g(t)|^7\right)^{\frac{1}{7}}$$ and notice that $$1=\max_{t \in [0,2]}\vert f(t)\vert=\max_{t \in [0,2]}\vert g(t)\vert=\max_{t \in [0,2]}\vert f(t)+g(t)\vert=\max_{t \in [0,2]}\vert f(t)-g(t)\vert$$
Hence
$$\Vert f+g \Vert^2+\Vert f-g \Vert^2 =2(1+2I)^2 \neq 4(1+I)^2= \left(\Vert f \Vert^2 + \Vert g \Vert^2\right)^2$$
as $I < \frac{1}{\sqrt 2}$.
Which allows to conclude.
Using the parallelogram identity is a general way to decide if a norm is induced by an inner product according to THE JORDAN-VON NEUMANN THEOREM