Which of the following determines an inner product over the space of real continuous functions with continuous first derivatives?

Question

First inner product is:

$\langle f,g\rangle = \int_0^1 f'(t)g'(t)dt + f(0)g(0)$

Second inner product is:

$\langle f,g\rangle = \int_0^1 f'(t)g'(t)dt$

I know that for something to be valid inner product it must have these properties

1. Hermitian symmetry $\langle x , y \rangle = \overline{ \langle x , y \rangle}$

2. Linearity1 $\langle sx, y \rangle = s \langle x, y \rangle$

3. Linearity2 $\langle x + y, z\rangle = \langle x, z \rangle + \langle y, z\rangle$

4. Positive definiteness $\langle x , x \rangle > 0$ if $x \neq 0$ and $\langle x, x\rangle = 0 \iff x = 0$

I wrote proof for the first one following all the rules but I thought it shouldn't so am I right or doing something wrong?

1. Recall $\bar \alpha = \alpha$ for $\alpha \in \mathbb R$:

$$ \begin{align} \langle f,g\rangle &= \int_0^1 f'(t)g'(t)dt + f(0)g(0) \\ &= \int_0^1 g'(t)f'(t)dt + g(0)f(0) \\ &= \langle g,f \rangle \end{align} $$

2.

$$ \begin{align} \langle sf,g\rangle &= \int_0^1 sf'(t)g'(t)dt + sf(0)g(0) \\ &= s\int_0^1 f'(t)g'(t)dt + f(0)g(0)\\ &= s\langle f,g \rangle \end{align} $$

3.

$$ \begin{align} \langle f + h,g\rangle &= \int_0^1 (f+ h)'(t)g'(t)dt + (f+h)(0)g(0) \\ &= \int_0^1 (f' + h')(t)g'(t)dt + (f+h)(0)g(0) \\ &= \int_0^1 \big(f'(t)g'(t)+h'(t)g'(t)\big)dt + f(0)g(0) + h(0)g(0) \\ &= \int_0^1 f'(t)g'(t)dt + \int_0^1 h'(t)g'(t)dt + f(0)g(0) + h(0)g(0) \\ &= \int_0^1 f'(t)g'(t)dt + f(0)g(0) + \int_0^1 h'(t)g'(t)dt + h(0)g(0) \\ &= \langle f ,g\rangle + \langle h,g\rangle \end{align} $$

4. Let 0 be the function with no support on [0,1].

$$ \begin{align} \langle 0, 0\rangle &= \int_0^1 0'(0)0'(0)dt + 0(0)0(0) \\ &= 0 \end{align} $$

And for any function that has support on [0,1] (and is smooth):

$$ \begin{align} \langle f, f\rangle &= \int_0^1 f'(0)f'(0)dt + f(0)f(0) \\ &= \int_0^1 f'(0)^2 + f(0)^2 \\ &> 0 \end{align} $$