Let's say we have $x\in [0,1]$ and we have the heat equation
$$\frac{d}{dt}u(x,t) = \Delta u(x,t) + f(u)$$ with $u(0,t) = u(1,t) = 0$ and $u(x,0)\in L^2$. $f$ is lipschitz continuous. I want to show local existence of the weak solution in $L^2$.
Here's what I did: $$u(x,t) = S(t)u_0 + \int_0^t S(t-s)f(u)ds$$
is the weak solution where $S(t)$ is the heat semigroup $(u_0 = u(x,0))$.
Define $$\mathcal{K}(v):= S(t)v_0 + \int_0^t S(t-s)f(v)ds$$
Then for $v, w\in L^2$ with $v_0 = w_0$ we consider
\begin{align} \sup_{t\in [0,T]}\left\| \mathcal{K}v - \mathcal{K}w \right\|_2 &= \sup_{t\in [0,T]}\left\|\int_0^T S(t-s)[f(v)-f(w)]\,ds\right\|_2\\ &\leq \sup_{t\in [0,T]}\int_0^T\left\|S(t-s)[f(v)-f(w)]\right\|_2\,ds\\ &\leq \sup_{t\in [0,T]}\int_0^T e^{-\omega_1(t-s)}C\left\|v-w\right\|_2 \,ds\\ &\leq T*1*C*\sup_{t\in [0,T]}\left\|v-w\right\|_2. \end{align}
In the last two steps, $C$ is the Lipschitz constant for $f$ and $\omega_1$ is the principal eigenvalue of the Laplace operator.
Pick $T$ small enough and we have a contraction mapping, so that implies the existence of a local solution. Does this argument work? Can I tweak it slightly to also work for $f$ locally Lipschitz? I was thinking for the locally Lipschitz case of maybe considering the ball in $L^2$ of radius $M$, and only choosing $v$ and $w$ in that ball.
We denote by $e^{t\Delta}f$ the convolution of $f$ with the heat kernel. As an example the heat Kernel in $\mathbb{R}$ is given by $$\Phi_t(x) = \frac{1}{\sqrt{4\pi t}} e^{-x^2/(4t)}$$ (a.k.a. the fundamental solution of the heat equation). If you take the Green's $G$ function for your problem (basically the fundamental solution with boundary conditions $G(0,t)=0=G(1,t)$) then this is your heat kernel for this problem. And $e^{t\Delta}f$ is just the convolution of $G$ with $f$ in $x$. What I want to say is that $e^{t\Delta}$ is your heat semigroup $S(t)$.
This heat semigroup has the following important property:
$$||e^{t\Delta} \phi||_q \leq (4\pi t)^{\frac{-1}{2}\left(\frac{1}{p}-\frac{1}{q}\right)} ||\phi||_p.$$
We are going to use this and the general Hölder inequality $$||fg||_r \leq ||f||_p ||g||_q,$$ where $1/r = 1/p + 1/q$ to construct a fixed-point argument in the following Banach space
$$X_\tau := \{u \in L^\infty((0,\tau);L^6([0,1]) : ||u||_\tau < \infty \},$$
where $$||u||_\tau := \sup_{0<t<\tau} t^{1/6} ||u(t)||_6.$$
Next we choose $M > ||u_0||_2$ and we denote by $B_{M,\tau}$ the closed ball in $X_\tau$ with center $0$ and radius $M$. We apply the fixed point argument to the following mapping $$S(v)(t) := e^{t\Delta}u_0 + \int_0^t e^{(t-s)\Delta}u^3(s) ds.$$
So let $u,v \in B_{M,\tau}$ we get \begin{align*} t^{1/6}||S(u)(t)-S(v)(t)||_6 &\leq t^{1/6} \int_0^t \left|\left|e^{(t-s)\Delta}\left(u^3(s)-v^3(s)\right)\right|\right|_6 ds \\ &\leq t^{1/6} \int_0^t (t-s)^{-1/6} \left|\left|u^3(s)-v^3(s)\right|\right|_2 ds, \end{align*} where we used the heat semigroup property. We estimate the norm term separately \begin{align*} \left|\left|u^3-v^3\right|\right|_2 &= \frac{1}{2} ||(u-v)(u^2+v^2)+(u+v)(u^2-v^2)||_2 \\ &\leq \frac{1}{2} \left(||u-v||_6 \left(||u^2||_3+||v^2||_3\right) + \left(||u||_6+||v||_6\right)||u^2-v^2||_3\right) \\ &\leq \frac{1}{2} \left(||u-v||_6 \left(||u^2||_3+||v^2||_3\right) + \left(||u||_6+||v||_6\right)||(u-v)(u+v)||_3\right) \\ &\leq \frac{1}{2} \left(||u-v||_6 \left(||u^2||_3+||v^2||_3\right) + \left(||u||_6+||v||_6\right)\left(||u||_6+||v||_6\right)||u-v||_6\right) \\ &\leq C \left(||u||_6^2 + ||v||_6^2\right) ||u-v||_6 \end{align*}
where we used Hölder's inequality and the triangel inequality a couple of times.
Putting everything together we get \begin{align*} t^{1/6}||S(u)(t)-S(v)(t)||_6 &\leq C t^{1/6} \int_0^t (t-s)^{-1/6} \left(||u||_6^2 + ||v||_6^2\right) ||u-v||_6 ds \\ &\leq C M^2 t^{1/6} \int_0^t (t-s)^{-1/6} s^{-2/6} ||u-v||_6 ds \\ &\leq C M^2 t^{1/6} \int_0^t (t-s)^{-1/6} s^{-1/2} ds \, \, ||u-v||_\tau \\ &\leq C M^2 \int_0^t s^{-1/2} ds \, \, ||u-v||_\tau \\ &\leq C M^2 \sqrt{t} ||u-v||_\tau \\ &\leq C M^2 \sqrt{\tau} ||u-v||_\tau \end{align*}
Note if we choose $\tau>0$ small enough,i.e., such that $CM^2 \sqrt{\tau} <1/2$, then we have a contraction mapping.
Next we look at the norm of $S(u)$
\begin{align*} t^{1/6} ||S(u)(t)||_6 &\leq t^{1/6} ||e^{t\Delta}u_0||_6 + CM^3\sqrt{\tau} \\ &\leq (4\pi)^{-1/6} ||u_0||_2 + CM^3\sqrt{\tau}, \end{align*}
where we used the heat semigropu property in the first term and the result from the difference estimate with $v=0$ for the second term. Note that $(4\pi)^{-1/6} < 1$, hence, we can choose $\tau$ small enough such that we have $$||S(v)||_\tau < M.$$
So $S$ is indeed a contraction mapping from $B_{M,\tau}$ to $B_{M,\tau}$. Thus, Banach's fixed-point theorem asserts the existence of unique integral solution in $B_{M,\tau}$.
Note that we have shown that $u(t) \in L^6$, therefore, $f(u) = u^3(t) \in L^2$. Since our unique fixed point satisfies $$u(t) = e^{t\Delta}u_0 + \int_0^t e^{(t-s)\Delta}f(u)(s) ds,$$ and using the smoothing properties of $e^{t\Delta}$, you should be able show that $u(t) \in L^2$ by using a bootstrap type argument.