I have alredy proved: $f, g$ two density functions. Prove $h(x)=$$\int_{-\infty}^{\infty} g(x-y)f(y) dy$ define a new density function. When $f$ and $g$ are $exp(\lambda)$ it's solved by $\int_{0}^{x} g(x-y)f(y)$ and the integral is solved easily.
Then is asked: $f, g$ are probability density functions of an normal distribution N(0,1), prove h is $N(0,\sqrt{2})$.
I tried: $g(x-y)f(y)=\frac{1}{2\pi}e^{\frac{-x^{2}+2xy-2y^{2}}{2}}$.
But i can't integrate that expresion and $N(0,\sqrt2)=\frac{1}{2\sqrt\pi}e^{\frac{-x^{2}}{4}}$ is such diferente form the other expression. For example, $\pi$ and $\sqrt\pi$
I assume by $N(0, \sqrt{2})$ you mean a normal variable with mean $0$ and variance $2$. Since $h(x)$ is the convolution of $f$ and $g$, and $f,g$ are the p.d.f of two independent normal variables, $h$ is the p.d.f of the sum of two standard normal variables, which has mean $0$ and variance $2$.
$h(x)=\int_{-\infty}^\infty g(x-y)f(y)dy$
$h(x)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\Big[ -\frac{(x-y)^2}{2}\Big] \frac{1}{\sqrt{2\pi}}\exp\Big[ -\frac{y^2}{2}\Big]dy$
$=\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sqrt{2 \pi}} \exp \Big[ -\frac{(x-y)^2+y^2}{2}\Big]dy$
$=\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sqrt{2 \pi}} \exp \Big[ -\frac{2y^2-2xy+x^2}{2}\Big]dy$
$=\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sqrt{2 \pi}} \exp \Big[ -\frac{y^2-2y\frac{x}{2}+\frac{x^2}{2}}{2 \frac{1}{2}}\Big]dy$
$=\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sqrt{2 \pi}} \exp \Big[ -\frac{y^2-2y\frac{x}{2}+\frac{x^2}{4}+\frac{x^2}{4}}{2 \frac{1}{2}}\Big]dy$
$=\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sqrt{2 \pi}} \exp \Big[ -\frac{(y-\frac{x}{2})^2+\frac{x^2}{4}}{2 \frac{1}{2}}\Big]dy$
taking all $x$ terms out of the integral, and multiplying the denominator by $\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{1}{\sqrt{2\pi} \sqrt{2}}\exp \Big[ -\frac{1}{2}\frac{x^2}{2}\Big]\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \frac{1}{\sqrt{2}}} \exp \Big[ -\frac{1}{2}\frac{(y-\frac{x}{2})^2}{ (\frac{1}{\sqrt{2}})^2}\Big]dy$
Now, recall that
$\frac{1}{\sqrt{2 \pi} \frac{1}{\sqrt{2}}} \exp \Big[ -\frac{1}{2}\frac{(y-\frac{x}{2})^2}{ (\frac{1}{\sqrt{2}})^2}\Big]$ is the p.d.f. of a normal variable $y$ with mean $\frac{x}{2}$ and standard deviation $\frac{1}{\sqrt{2}}$, therefore its integral from $-\infty$ to $\infty$ should be one for every $x$. Which means
$h(x)=\frac{1}{\sqrt{2\pi} \sqrt{2}}\exp \Big[ -\frac{1}{2}(\frac{x}{\sqrt{2}})^2\Big]$ which is the pdf of a normal variable with mean $0$, and variance $2$.
Sorry for keeping the improper fraction $\frac{1}{\sqrt{2}}$, as it was for illustrative purposes.