This is 1.25 on Brezis Functional analysis.
Define the semiscalar product [x,y] by $$[x,y]=\inf_{t>0}\frac{1}{2t}[||x+ty||^2-||x||^2].$$ E be an n.v.s. I donot know how to prove that
1.$[x,\lambda x+\mu y]=\lambda||x||^2+\mu[x,y]\ \forall x,y\in E,\lambda\in\mathbb{R},\forall \mu >0.$
2.$[\lambda x,\mu y]=\lambda\mu[x,y],\forall x,y\in E,\forall\lambda,\mu\geq 0$
I probably understand that this semiscalar product is similar to an inner product, but I only have trigonometric inequalities for its norm, how do I prove that this product is linear?
This is currently a partial answer as I'll only show $2.$, because I don't know know how to prove $1.$ yet:
Let $\lambda, \mu > 0$, and $(x,y) \in E^2$. Then, we have: $$\begin{split}[\lambda x, \mu y] &= \inf_{t > 0} \frac{1}{2t} \left(\|\lambda x + t\mu y\|^2 - \|\lambda x\|^2\right)\\ &= \inf_{t' > 0} \frac{1}{2\lambda t'} \left(\|\lambda x + \lambda t'\mu y\|^2 - \|\lambda x\|^2\right) \quad \text{where } t' := \frac{t}{\lambda}\\ &= \inf_{t' > 0} \frac{\lambda}{2t'} \left(\|x + t'\mu y\|^2 - \| x\|^2\right)\\ &= \inf_{t'' > 0} \frac{\lambda \mu}{2t''} \left(\|x + t'' y\|^2 - \|x\|^2\right) \quad \text{where } t'' := \mu t'\\ &= \lambda \mu [x,y]\end{split}$$ Since the cases $\lambda = 0$ and $\mu = 0$ are trivially true, we're done with $2.$.
To put it into words, I simply used the fact that the whole of $(0, \infty)$ is its own image through the multiplication by any positive real, thus the infimum will be the same whether we dilate or expand the variable $t$ by a certain positive factor.
To compensate for not being able to show $1.$, let me explain how you can get $[x, \lambda x] = \lambda \|x\|^2$ for $\lambda \in \mathbb{R}$:
We can see that by $2.$ we only need to calculate $[x,x]$ and $[x, -x]$.
Let's handle them simultaneously: $$[x, \pm x] = \inf_{t > 0} \frac1{2t}\left(\|x \pm t x\|^2 - \|x\|^2\right) = \inf_{t > 0} \frac{(1 \pm t)^2 - 1}{2t}\|x\|^2 = \inf_{t > 0} \left(\pm 1 + \frac{t}{2}\right)\|x\|^2 = \pm \|x\|^2$$
EDIT: now that I know that it came from Brezis' Functional Analysis, I was able to check the exercise this came from. I realised that my edition didn't have any of the exercises for some reason so I had to go fetch an online copy but oh well. The edition I have available states this as item $3$ not item $2$ but it could just be a difference in editions.
Anyway, the first item states, when applied to $[\cdot, \cdot]$, that: $$[x,y] = \lim_{t \to 0^+} \frac{1}{2t}\left(\|x+ty\|^2 - \|x\|^2\right) \in [-\infty, +\infty)$$ while the second item, which reads $|[x,y]| \leq \|x\| \|y\|$, lets us know that $[x,y]$ is in fact finite.
Being able to write this with limits makes this much simpler than with $\inf$. We have, for $\lambda \neq 0$: $$\begin{split}[x,\lambda x + \mu y] - \lambda \|x\|^2 &= \lim_{t \to 0^+} \frac{1}{2t}\left(\|x+t(\lambda x + \mu y)\|^2 - \|x\|^2\right) - \lim_{t \to 0^+} \frac{1}{2t}\left(\|x+t\lambda x\|^2 - \|x\|^2\right)\\ &= \lim_{t \to 0^+} \left(\frac{1}{2t}\left(\|x+t(\lambda x + \mu y)\|^2 - \|x\|^2\right) - \frac{1}{2t}\left(\|x + t\lambda x\|^2 - \|x\|^2\right)\right)\\ &= \lim_{t \to 0^+} \frac{1}{2t}\left(\|x+t(\lambda x + \mu y)\|^2 - \|x + t\lambda x\|^2\right)\\ &= \lim_{t \to 0^+ \atop t < \frac{1}{|\lambda|}} \frac{1}{2t}\left(\left\|(1+t\lambda) x + t \frac{1 + \lambda t}{1 + \lambda t}\mu y\right\|^2 - \|(1 + t\lambda) x\|^2\right)\\ &= \lim_{t \to 0^+ \atop t < \frac{1}{|\lambda|}} \frac{(1 + \lambda t)^2}{2t}\left(\left\|x + \frac{t}{1 + \lambda t}\mu y\right\|^2 - \|x\|^2\right)\\ &= \lim_{t \to 0^+ \atop t < \frac{1}{|\lambda|}} \frac{(1 + \lambda t) + (\lambda t + \lambda^2 t^2)}{2t}\left(\left\|x + \frac{t}{1 + \lambda t}\mu y\right\|^2 - \|x\|^2\right)\end{split}$$ where I only included the $t < \frac{1}{|\lambda|}$ to emphasize that there's no problem with writing $\frac{1}{1 + \lambda t}$ in that case and taking the limit (there could have been a problem if the denominator was something more complicated which had a sequence of zeros $t_n$ tending themselves to $0$ when $n \to \infty$).
Now, it should be clear that $\|x+s\mu y\|^2 \to \|x\|^2$ when $s \to 0$ by continuity of the norm, and we have $\frac{t}{1 + \lambda t} \underset{t \to 0}{\sim} t \to 0$ when $t \to 0$, hence: $$\frac{\lambda t + \lambda^2 t^2}{2t}\left(\left\|x + \frac{t}{1 + \lambda t}\mu y\right\|^2 - \|x\|^2\right) = \frac{\lambda + \lambda^2 t}{2}\left(\left\|x + \frac{t}{1 + \lambda t}\mu y\right\|^2 - \|x\|^2\right) \xrightarrow[t \to 0^+]{} 0$$
To conclude, note that we already know that $\frac{1}{2s}\left(\|x+s\mu y\|^2 - \|x\|^2\right)$ converges to $[x, \mu y]$ when $s \to 0^+$, and as already stated $\frac{t}{1 + \lambda t}$ tends to $0$ when $t \to 0$, thus: $$ [x,\lambda x + \mu y] - \lambda \|x\|^2 = \lim_{t \to 0^+ \atop t < \frac{1}{|\lambda|}} \frac{1}{2\frac{t}{1 + \lambda t}}\left(\left\|x + \frac{t}{1 + \lambda t}\mu y\right\|^2 - \|x\|^2\right) = [x, \mu y]$$
There could of course exist a much simpler proof but I hope this one works for you.