Upsampling signal using convolution-based interpolation filters
Starting from the equation with the convolutions and assuming unit distance between sampling grid points, for a one-dimensional signal you'll have just:

$$I(\mathbf{x}) = I((x_1))= I_s((p_1))*h(x_1),$$

where $\mathbf{x}$ is the vector of continuous coordinates of which there is only one: $x_1 \in \mathbb{R}.$ Likewise there is just a single discrete coordinate $p_1 \in \mathbb{Z}.$ For simplicity, you could drop the vector notation and subscripts altogether and have just:

$$I(x) = I_s(p)*h(x).$$

Without having fully read the paper, I don't think the authors dare to venture into defining what is meant by convolution between discrete-time $I_s(p)$ and continuous-time $h(x)$ which results in continuous-time $I(x)$. Such a convolution can be defined as:

$$I(x) = \sum_{p=-\infty}^{\infty}I_s(p)h(x-p)$$

For example, if you want to obtain $I(x)$ within the inter-sample interval $0 \le x &lt; 1$ and you use the linear interpolation kernel, then the above simplifies as:

$$\begin{eqnarray}I(x) &amp;=&amp; \sum_{p=-\infty}^{\infty}I_s(p)h_{Lin}(x-p)\\
&amp;=&amp; \sum_{p=-\infty}^{\infty}I_s(p)\times\cases{
1-|x-p|, &amp;if$\quad 0 \leqslant |x-p| &lt; 1,$\\
0, &amp;if$\quad 1 \leqslant |x-p|$
}\\
&amp;=&amp; \sum_{p=0}^{1}I_s(p)(1-|x-p|)\\
&amp;=&amp; I_s(0)(1-x) + I_s(1)(1-(1-x))\\
&amp;=&amp; I_s(0)(1-x) + I_s(1)x.\end{eqnarray}$$

Or likewise for any inter-sample interval $x_0 \le x &lt; x_0 + 1:$

$$I(x) = I_s(x_0)(1-(x - x_0)) + I_s(x_0 + 1)(x - x_0).$$

Another way to understand such convolutions is that $I_s(p)$ is the product of the continuous-time signal multiplied by a Dirac comb, and to use a continuous-time convolution integral.

As you can see from the above, the interpolation filter is not a regular discrete-time filter, but one that converts a discrete-time signal into a continuous-time signal. The convolution filter definitions cited in the question are directly applicable, without further "sizing", when the sampling interval is 1.