Calculate the norm of the functional $f(x)=\displaystyle\int_{-1}^1tx(t)dt$ Find the norm of the functional $f(x)=\displaystyle\int_{-1}^1tx(t)dt$ in the space $C^1[-1,1]$, where the norm is given by $\|x\|=\max\limits_{t\in[-1,1]}|x(t)|+\max\limits_{t\in[-1,1]}|x'(t)|$. The best upper bound I managed to get: integrating by parts, we get $$f(x)=\displaystyle\int_{-1}^1\left(\dfrac{1}{2}-\dfrac{t^2}{2}\right)x'(t)dt.$$
Adding this to the original functional and dividing by two, we get: $$f(x)=\displaystyle\int
_{-1}^1\left(\dfrac{t}{2}x(t)+\left(\dfrac{1}{4}-\dfrac{t^2}{4}\right)x'(t)\right)dt,$$
so
$$|f(x)|\le\int_{-1}^1\left(\dfrac{|t|}{2}|x(t)|+\left|\dfrac{1}{4}-\dfrac{t^2}{4}\right||x'(t)|\right)dt\le\|x\|\cdot\int_{-1}^1\max\left\{\dfrac{|t|}{2},\left|\dfrac{1}{4}-\dfrac{t^2}{4}\right|\right\}dt=$$
$$=\frac{2\sqrt2-1}{3}\cdot\|x\|\approx0,609476\cdot\|x\|.$$
However, I can't find even numerical lower bound for $0,609476$ using $\dfrac{|f(x)|}{\|x\|}$ fractions. I have a feeling that received upper bound is overstated. I would be grateful for any comments and ideas, both on how to improve the upper bound, and how to get at least a numerical lower bound.
Interestingly, the norm of the same functional under the condition that $\|x\|=\max\left\{\max\limits_{t\in[-1,1]}|x(t)|,\max\limits_{t\in[-1,1]}|x'(t)|\right\}$ is calculated trivially and is equal to $\dfrac{2}{3}$.

Thanks to Brifa's comment, we managed to improve the upper bound to $\|f\|\le0,4$. For a while I thought that the norm is $\frac{1}{3}$, but found the following example (see graph):

The "switch" point is approximately $0.806$, and $\dfrac{|f(x)|}{\|x\|}\approx0,35$.
 A: First, let us consider the decomposition of $x$ into even and odd parts, $x = x_e + x_o$.  Then clearly, $f(x) = f(x_o)$.  On the other hand, since $x_o(t) = \frac{1}{2} [f(t) - f(-t)]$, we see that $\sup_{t\in [-1, 1]} |x_o(t)| \le \sup_{t\in [-1, 1]} |x(t)|$, and $\sup_{t\in [-1, 1]} |x_o'(t)| \le \sup_{t\in [-1, 1]} |x'(t)|$, so $\lVert x_o \rVert \le \lVert x \rVert$.  Therefore, if $x_o$ is not identically zero, then $\frac{|f(x_o)|}{\lVert x_o \rVert} \ge \frac{|f(x)|}{\lVert x \rVert}$.  From this (and the fact that if $x_o = 0$ then $f(x) = 0$), we see that the norm of $f$ is the same as the norm of $f$ restricted to odd $x$.
Now, restricting further to odd $x$ with $\lVert x \rVert = 1$, let us suppose $\operatorname{sup}_t |x(t)| = a$ and $\operatorname{sup}_t |x'(t)| = 1-a$.  Then first of all, since $|x(t)| \le 1-a$ for all $t\in [-1, 1]$ using the mean value theorem, we must have $a \le \frac{1}{2}$.  Now whenever $t\ge 0$, we must have $x(t) \le \operatorname{min}((1-a)t, a)$, and whenever $t\le 0$, we must have $x(t) \ge \operatorname{max}((1-a)t, -a)$.  From this, we can conclude
$$f(x) \le \int_{-1}^{-a/(1-a)} -at\,dt + \int_{-a/(1-a)}^{a/(1-a)} (1-a)t^2\,dt + \int_{a/(1-a)}^1 at\,dt = \frac{2a^3-6a^2+3a}{3(1-a)^2}.$$
It is straightforward to see that this is maximized when $2a^3 - 6a^2 + 9a - 3 = 0$.  Numerically, the root of this polynomial in the interval $[0, 1/2]$ is approximately $0.446$ and the maximum value is approximately $0.350$.  On the other hand, we also see $f(x) = -f(-x) \ge -\sup_{a\in [0,1/2]} \frac{2a^3 - 6a^2 + 3a}{6(1-a)^2}$.  Therefore, the norm of $f$ is at most $\sup_{a\in [0,1/2]} \frac{2a^3 - 6a^2 + 3a}{3(1-a)^2}$.
On the other hand, if we take the piecewise linear function suggested by the previous paragraph, and then take $C^1$ joins of the pieces which do not affect $f$ much while preserving the norm of $x$, then we can get $C^1$ functions $x$ with norm 1 such that $f(x)$ is arbitrarily close to $\sup_{a\in [0, 1/2]} \frac{2a^3 - 6a^2 + 3a}{3(1-a)^2}$.  Therefore,
$$\lVert f \rVert = \sup_{a\in [0,1/2]} \frac{2a^3 - 6a^2 + 3a}{3(1-a)^2} \approx 0.350.$$
