$L^2$ is the only Hilbert space , Parallelogram Law and particular $f(t),g(t)$ Let the space $C([0,1])$ and consider the norm $\forall p \in \mathbb{N}$
$$\forall f \in  C([0,1]),
   ||f ||_{L^P}= \left (  \int^1_0 |f(t)|^P dt \right )^{\frac{1}{p}} $$
knowing that when $p=2$, this norm is the norm induced by the innder product 
$$\forall f,g \in C([0,1]), <f,g>=\int^1_0 f(t)\overline{g(t)} dx $$
The goal of this excersice is to prove that if $p \neq 2$ $||.||_p$ is not a norm induced by an inner product (i.e $L^2$ is the only hilbert space among all $L^p$ spaces. )
To do so, study when the parallelogram law holds. Hint: consider the functions 
$$f(t)=\frac{1}{2}-t; 
 g(t)= 
  \begin{matrix} 
     \frac{1}{2} -t & \text{ if } 0 \leq t \leq \frac{1}{2} 
     \\ t-\frac{1}{2} & \text{ if  } \frac{1}{2} <t \leq 1 \end{matrix} $$

Parallelogram Law $M=C([0,1])$
Let $(M,<.,.>)$ is an inner product space, where induced norm 
$$|| .|| =\sqrt{<.,.>} $$
then 
$$\forall x,y \in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2) $$

Using u-sub $u=(1/2 -t)$ and $\frac{du}{dt}=-1 $ so $dt=-du$
$$ \begin{aligned} 
||f|| &= \int^1_0 (|f(t)|^p dt)^{1/p}= \int^1_0 (|1/2 -t|^p dt)^{1/p}
   \\ &=\int^{u(1)}_{u(0)} (u^p -du)^{1/p}
        \\& =(\int^{-1/2}_{1/2} u^p du)^{1/p}
         \\ &= \left [ \left (\frac{u^{p+1}}{p+1}  \right )^{1/p} \right]^{-1/2}_{1/2}
        \\&= \left [\frac{u^{1+1/p}}{(p+1)^{1/p}} \right]^{-1/2}_{1/2}
        \\&=\frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}}
\end{aligned}$$

working on $||f-g ||,||f+g||,||g||$ 

Guessing big picture that the parallelogram law only works when $p=2$ so it is only induced norm when $p=2$
 A: Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<\infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $p\ne q$ these are not isomorphic.
A: Easier (and slightly more careful) proof.
First: Show that $L^2 [ 0,1]$ as a normed linear space over the reals $\mathbf{R}$ is complete. There are several ways of showing this; one way which quickly comes to my mind is to show that Cauchy in $L^2$ norm implies Cauchy in $L^1$ norm (Jensen's inequality would work), and that $L^2 [ 0,1]$ is a closed subspace of $L^1[0,1]$ (monotone convergence theorem is one sledgehammer you can use to crack this nut). This will tell you that to show that $L^2[0,1]$ is a Hilbert space, it now suffices to show that $|| \cdot ||_{L^2}$ is induced by an inner product.
Note: It is not true that all inner product spaces are Hilbert spaces; by definition, in a Hilbert space all Cauchy sequences should converge.
Now, coming to the main business: checking that $|| \cdot ||_{L^2}$ satisfies the parallelogram identity suffices. In this direction, you can just do this vertification by expanding squares using the identity $(a+b)^2 = a^2 + 2ab + b^2.$
It would be fun to show that $L^p [0,1]$ is not a Hilbert space if $1 \le p \le \infty$ and $p \ne 2.$
Take the functions $f (t) = \mathbf{1} [ t \in [0,1/2)]$ and $g(t) = \mathbf{1} [ t \in [1/2 , 1].$ Then $||f||_p^2 = 1/2^{2/p} = ||g||_p^2;$ and $||f + g||_p^2 = 1/2 = ||f-g||_p^2,$ so for the parallelogram identity to be satisfied one must have $\frac{1}{2^{2/p}} + \frac{1}{2^{2/p}} = 1,$ and clearly this is true if and only if $p=1/2;$ thus for $p \ne 1/2$ the space $L^p [ 0,1]$ is not even an inner product space (in the sense that the corresponding norm is not induced by an inner product), let alone the question of it being a Hilbert space.
As pointed out by grew in the comments, Adam's solution requires some extra work, but it is otherwise very nice.
