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The following function $f(t,x):[0,T]\times R\mapsto R$ such that $\int^T_0|f(t,0)|^2 d t<\infty$, where $0<T<\infty$.

If $f(t,x)$ satisfies $|f(t,x)|\leq Ax+B$ for each $x\in R$ and $A, B$ are given constants, then we say that $f(t,x)$ is of linear growth in $x$. All Lipschitz functions are of linear growth.

If $f(t,x)$ satisfies $|f(t,x)|\leq C+C|x|^p$ for each $x\in R$, $C>0$ and $p\geq1$ are given constants, then we say $f(t,x)$ is of polynominal growth in $x$.

Besides, if there exists a strictly increasing convex function $\varphi(\cdot)$ with $\varphi(0)=\varphi(0+)$ such that for each $x\in R$, $|f(t,x)|\leq g_t+\varphi(x)$, where $g_t$ is suqare integrable with respect to $t$, then we say $f(t,x)$ is convex growth in $x$. Convex growth functions contain many types of functions, such as exponential functions, tangent functions, even linear growth functions and polynominal growth functions.

Yet there exists a general growth. We say $f(t,x)$ has a general growth if for each $\alpha\geq 0$, $\sup_{|x|\leq \alpha}|f(t,x)-f(t,0)|\in L^1([0,T])$, where $L^1$ means Lebesgue integrable.

I am very confused that from theoretical point, the general growth is actually more general than the convex growth, but I can not find an explicit function satisfies the general growth other than the convex growth. All functions I can find can be controlled by a convex function, i.e., they are all convex growth.

Anyone can help me to find a function with general growth but not convex growth?

Here is my example: $e^{-xt^2}$. I do not know if it is right.

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  • $\begingroup$ Something's not right. For each fixed $\alpha$, $\sup_{|x|\le \alpha} |f(x) - f(0)|$ is a number, not a function of anything, so what is the variable with respect to which it is supposed to be $L^1$? Do you mean there is a locally $L^1$ function $g(\alpha)$ such that $\sup_{|x|\le \alpha} |f(x) - f(0)| \le g(\alpha)$? It can't be globally $L^1$ if you allow any growth at all. $\endgroup$ – Robert Israel Aug 29 '15 at 1:42
  • $\begingroup$ @RobertIsrael Sorry, my mistake. Another variable, time t, should be considered. $\endgroup$ – XIAO Lishun Aug 29 '15 at 2:08
  • $\begingroup$ Your example is bounded by $1$. $\endgroup$ – Robert Israel Aug 30 '15 at 4:08
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You might try $f(t,x) = t^{-1/2} x^2$ for $t > 0$, $f(0,x) = 0$.

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