Let $C^1[0,1]$ the space of all the continuous functions $[0,1]\Rightarrow \mathbb{C}$ with continuous derivative on $[0,1]$ which are the following is an inner product
$$\langle f,g\rangle = f(0) \overline{g(0)} + \int_0^1 f'(t) \overline{g'(t)} \, dt$$
$$\langle f,g\rangle =f(0) \overline{g(0)}+f'(1)\overline{g'(1)} \, dt$$
$$\langle f,g\rangle =2\int_0^{\frac{1}{2}} f(t) \overline{g(t)} \, dt-\int_{\frac{1}{2}}^1 f(t) \overline{g(t)} \, dt$$
- a. $$\overline{<g,f>}=\overline{g(0)\overline{f(0)}+\int_{0}^{1}g'(t)\overline{f'(t)}dt}=f(0)\overline{g(0)}+\int_{0}^{1}f'(t)\overline{g'(t)}dt=<f,g>$$
b.$$<\alpha f+\beta g,h>=[\alpha f(0)+\beta g(0)]\overline{h(0)}+\int_{0}^{1}[\alpha f'(t)+\beta g'(t)]\overline{h'(t)}dt=\alpha f(0)\overline{h(0)}+\alpha\int_{0}^{1}f'(t)\overline{h'(t)}dt+\beta g(0)\overline{h(0)}+\beta \int_{0}^{1} g'(t)\overline{h'(t)}dt=\alpha<f,h>+\beta<g,h>$$
c.$$<f,f>=|f(0)|^2+\int_{0}^{1}|f'(t)|^2dt>0$$ and $$|f(0)|^2+\int_{0}^{1}|f'(t)|^2dt=0\iff |f(0)|^2=-\int_{0}^{1}|f'(t)|dt\iff f\equiv 0$$
Therefore $<f,g>=f(0)\overline{g(0)}+\int_{0}^{1}f'(t)\overline{g'(t)}dt$ is an inner product
2.
For $f=x(x-1)^2$ $<f,f>=0$ but $f\not\equiv0$ so $<f,g>=f(0)\overline{g(0)}+f'(1)\overline{g'(1)}dt$ is not an inner product
- a. $$\overline{<g,f>}=\overline{2\int_{0}^{\frac{1}{2}}g(t)\overline{f(t)}dt-\int_{\frac{1}{2}}^{1}g(t)\overline{f(t)}dt}=2\int_{0}^{\frac{1}{2}}f(t)\overline{g(t)}dt-\int_{\frac{1}{2}}^{1}f(t)\overline{g(t)}dt=<f,g>$$
b. $$<\alpha f+\beta g, h>=2\int_{0}^{\frac{1}{2}}[\alpha f(t)+\beta g(t)]\overline{h(t)}dt-\int_{\frac{1}{2}}^{1}[\alpha f(t)+\beta g(t)]\overline{h(t)}dt=2\int_{0}^{\frac{1}{2}}\alpha f(t)\overline{h(t)}dt-\int_{\frac{1}{2}}^{1}\alpha f(t)\overline{h(t)}dt+2\int_{0}^{\frac{1}{2}}\beta g(t)\overline{h(t)}dt-\int_{\frac{1}{2}}^{1}\beta g(t)\overline{h(t)}dt=\alpha<f,h>+\beta<g,h>$$
c. $$<f,f>=2\int_{0}^{\frac{1}{2}}|f(t)|^2-\int_{\frac{1}{2}}^{1}|f(t)|^2\geq 0$$
For it to be equal to $0$ we need $2\int_{0}^{\frac{1}{2}}|f(t)|^2=\int_{\frac{1}{2}}^{1}|f(t)|^2$ and it seems that we can find sone exponent function that will fulfil it, but how can we prove it? do 1 and 2 are correct?
For 3.: Assume $\langle \,,\rangle$ is an inner product.
Let $f = 0$ on $[0,1/2],$ $f(x) = (x-1/2)^2$ on $[1/2,1].$ Then $f\in C^1([0,1])$ and $f\not \equiv 0.$ However,
$$\tag 1\langle f,f \rangle = 2\int_0^{1/2}|f|^2 -\int_{1/2}^1 |f|^2 = 2\cdot 0 -\int_{1/2}^1 |f|^2= -\int_{1/2}^1 |f|^2 .$$
Because $\langle \,,\rangle$ is an inner product, the left side of $(1)$ is positive. But the right side is negative. That is a contradiction, and therefore $\langle \,,\rangle$ is not an inner product.
Your proofs in 1,2 are pretty good, especially 2. On your proof in 1.c: You have the right idea, but I might write it like this: First it's clear that $\langle f,f \rangle \ge 0$ for all $f\in C^1[0,1].$ Suppose $\langle f,f \rangle = 0.$ Then both $f(0)=0$ and $\int_0^1 (f')^2 = 0.$ Because $(f')^2$ is continuous and nonnegative, we have $f'\equiv 0.$ Thus $f$ is constant, hence $f\equiv f(0) = 0.$