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Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} \int_{[-1,1]} (\rho(.,t)*K_{T-t})(x) dxdt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $t$. So $f_T$ is the temperature at time $T$.

Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} (\rho(.,t)*K_{T-t})(x) dt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $t$. So $f_T$ is the temperature at time $T$.

Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} \int_{[-1,1]} (\rho(.,t)*K_{T-t})(x) dxdt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $T$. So $f_T$ is the temperature at time $t$.

Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} \int_{[-1,1]} (\rho(.,t)*K_{T-t})(x) dxdt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $t$. So $f_T$ is the temperature at time $T$.

Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} \int_{[-1,1]} (\rho(.,t)*K_{T-t})(x) dxdt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $t$. So $f_T$ is the temperature at time $t$.

Suppose we have a function $\rho(x,t)$ defined over $[-1,1]\times[0,1]$. Define the integral $ f_T(x)=\int_{[0,1]} \int_{[-1,1]} (\rho(.,t)*K_{T-t})(x) dxdt $ for $x\in R$ and $T>1$, where $*$ is the convolution, and $K_a$ is the Gaussian kernel with variance $a$.

The question is given $f_T$ for some $T>1$, can we determine $\rho$ uniquely (noting that $\rho$ has a bounded support)?

$f_T$ has the following intuitive meaning: Suppose we have an ideal metal rod of infinite length . $\rho(x,t)$ is the rate of heat flow into $x$ at time $T$. So $f_T$ is the temperature at time $t$.