Consider this answer as a supplement material to the other answers provided.
The upsampling method (mean point between every two points) we are talking about can be represented in the form of equations as described below.
Let $x[n]$ be the original signal and $x_{up}[n]$ be the upsampled version of $x[n]$ obtained by linear interpolation technique. The equation presented below establishes the relationship between $x_{up}[n]$ and $x[n]$.
\begin{equation}
x_{up}[2n] = x[n] \label{1}
\end{equation}
\begin{equation}
x_{up}[2n + 1] = \frac{x[n] + x[n + 1]}{2}
\end{equation}
To understand the performance of this method let's apply this method on some fundamental signals and examine the resultant up-sampled signal.
1. Kronecker-Delta signal:
Let the upsampled version of the Kronecker-delta function ${\delta}[n]$, obtained by using the linear interpolation method described above be ${\delta}_{up}[n]$. Then it can be seen that ${\delta}_{up}[-1]$ and ${\delta}_{up}[1]$ have non-zero values. So the upsampled signal ${\delta}_{up}[n]$ is a combination of 3 Kronecker-delta (with delay) as shown below:
\begin{equation}
{\delta}_{up}[n] = {\delta}[n] + 0.5 {\delta}[n + 1] + 0.5 {\delta}[n - 1]
\end{equation}
Thus the distortion (let's call it $d_{up}[n]$) introduced by the linear interpolation techniques on the Kronecker-delta function can be quantified as
\begin{equation}
d_{up}[n] = 0.5 {\delta}[n + 1] + 0.5 {\delta}[n - 1]
\end{equation}
The energy of distortion is of the order of the signal in this case.
2. Unit step signal:
Let the upsampled version of the unit step signal $u[n]$, obtained by using the linear interpolation method described above be $u_{up}[n]$. Then it can be seen that $u_{up}[n]$ has non-zero for $n = -1$. So the upsampled signal $u_{up}[n]$ is a combination of unit-step function and time delayed versions of unit step as shown below:
\begin{equation}
u_{up}[n] = u[n] + \frac{ {\delta}[n + 1]}{2}
\end{equation}
Thus the distortion (let's call it $d_{up}[n]$) introduced by the linear interpolation techniques on the unit-step function can be quantified as
\begin{equation}
d_{up}[n] = \frac{ {\delta}[n + 1]}{2}
\end{equation}
The energy of distortion is arguably infinitesimally small in comparison with the energy of the signal in this case.
3. Sine Tone signal:
Let $s[n]$ be a pure sine tone signal as described below
$$s[n] = \sin\bigg(\frac{2 \pi n} {N}\bigg)$$
Let $s_{up}[n]$ be the signal obtained by applying linear interpolation on the signal $s[n]$
Then
\begin{equation}
\begin{split}
s_{up}[2n + 1] &= \frac{(s[n] + s[n + 1])}{2} \\
& = \frac{\bigg(\sin\bigg(\frac{2 \pi n} {N}\bigg) + \sin\bigg(\frac{2 \pi (n + 1)} {N}\bigg)\bigg)}{2} \\
\end{split}
\end{equation}
Simplifying the above expression using reduced form of the $sinA + sinB$ we get the below
\begin{equation}
s_{up}[2n + 1] = \sin\bigg(\frac{2 \pi (2 n + 1)} {2 N}\bigg) \cos\bigg(\frac{2 \pi}{2 N}\bigg)
\end{equation}
The expected value at point $(2n + 1)$ in the upsampled signal is $\sin\bigg(\frac{2 \pi (2n + 1)} {2 N}\bigg)$
Hence the distortion (let's call it $d_{up}[n]$) introduced in the signal is:
\begin{equation}
d_{up}[2n + 1] = \sin\bigg(\frac{2 \pi (2 n + 1)} {2 N}\bigg) \bigg( 1 - \cos\bigg(\frac{2 \pi}{2 N}\bigg)\bigg)
\end{equation}
The magnitude of distortion depends on the value of $N$